Calculus Examples

$s (t) = \frac{25000 e^{- t}}{{(1 + 5 e^{- t})}^{2}}$

Step 1

Find the first derivative of the function.

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Step 1.1

Since $25000$ is constant with respect to $t$ , the derivative of $\frac{25000 e^{- t}}{{(1 + 5 e^{- t})}^{2}}$ with respect to $t$ is $25000 \frac{d}{d t} [\frac{e^{- t}}{{(1 + 5 e^{- t})}^{2}}]$ .

$25000 \frac{d}{d t} [\frac{e^{- t}}{{(1 + 5 e^{- t})}^{2}}]$

Step 1.2

Differentiate using the Quotient Rule which states that $\frac{d}{d t} [\frac{f (t)}{g (t)}]$ is $\frac{g (t) \frac{d}{d t} [f (t)] - f (t) \frac{d}{d t} [g (t)]}{{g (t)}^{2}}$ where $f (t) = e^{- t}$ and $g (t) = {(1 + 5 e^{- t})}^{2}$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} \frac{d}{d t} [e^{- t}] - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{({(1 + 5 e^{- t})}^{2})}^{2}}$

Step 1.3

Multiply the exponents in ${({(1 + 5 e^{- t})}^{2})}^{2}$ .

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Step 1.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} \frac{d}{d t} [e^{- t}] - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{2 \cdot 2}}$

Step 1.3.2

Multiply $2$ by $2$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} \frac{d}{d t} [e^{- t}] - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 1.4

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - t$ .

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Step 1.4.1

To apply the Chain Rule, set $u_{1}$ as $- t$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d t} [- t]) - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 1.4.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} (e^{u_{1}} \frac{d}{d t} [- t]) - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 1.4.3

Replace all occurrences of $u_{1}$ with $- t$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} (e^{- t} \frac{d}{d t} [- t]) - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 1.5

Differentiate.

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Step 1.5.1

Since $- 1$ is constant with respect to $t$ , the derivative of $- t$ with respect to $t$ is $- \frac{d}{d t} [t]$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} e^{- t} (- \frac{d}{d t} [t]) - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 1.5.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} e^{- t} (- 1 \cdot 1) - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 1.5.3

Simplify the expression.

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Step 1.5.3.1

Multiply $- 1$ by $1$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} e^{- t} \cdot - 1 - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 1.5.3.2

Move $- 1$ to the left of ${(1 + 5 e^{- t})}^{2} e^{- t}$ .

$25000 \frac{- 1 \cdot ({(1 + 5 e^{- t})}^{2} e^{- t}) - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 1.5.3.3

Rewrite $- 1 {(1 + 5 e^{- t})}^{2}$ as $- {(1 + 5 e^{- t})}^{2}$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 1.6

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{2}$ and $g (t) = 1 + 5 e^{- t}$ .

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Step 1.6.1

To apply the Chain Rule, set $u_{2}$ as $1 + 5 e^{- t}$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - e^{- t} (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d t} [1 + 5 e^{- t}])}{{(1 + 5 e^{- t})}^{4}}$

Step 1.6.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 2$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - e^{- t} (2 u_{2} \frac{d}{d t} [1 + 5 e^{- t}])}{{(1 + 5 e^{- t})}^{4}}$

Step 1.6.3

Replace all occurrences of $u_{2}$ with $1 + 5 e^{- t}$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - e^{- t} (2 (1 + 5 e^{- t}) \frac{d}{d t} [1 + 5 e^{- t}])}{{(1 + 5 e^{- t})}^{4}}$

Step 1.7

Differentiate.

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Step 1.7.1

Multiply $2$ by $- 1$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 2 e^{- t} ((1 + 5 e^{- t}) \frac{d}{d t} [1 + 5 e^{- t}])}{{(1 + 5 e^{- t})}^{4}}$

Step 1.7.2

By the Sum Rule, the derivative of $1 + 5 e^{- t}$ with respect to $t$ is $\frac{d}{d t} [1] + \frac{d}{d t} [5 e^{- t}]$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 2 e^{- t} ((1 + 5 e^{- t}) (\frac{d}{d t} [1] + \frac{d}{d t} [5 e^{- t}]))}{{(1 + 5 e^{- t})}^{4}}$

Step 1.7.3

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 2 e^{- t} ((1 + 5 e^{- t}) (0 + \frac{d}{d t} [5 e^{- t}]))}{{(1 + 5 e^{- t})}^{4}}$

Step 1.7.4

Add $0$ and $\frac{d}{d t} [5 e^{- t}]$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 2 e^{- t} ((1 + 5 e^{- t}) \frac{d}{d t} [5 e^{- t}])}{{(1 + 5 e^{- t})}^{4}}$

Step 1.7.5

Since $5$ is constant with respect to $t$ , the derivative of $5 e^{- t}$ with respect to $t$ is $5 \frac{d}{d t} [e^{- t}]$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 2 e^{- t} ((1 + 5 e^{- t}) (5 \frac{d}{d t} [e^{- t}]))}{{(1 + 5 e^{- t})}^{4}}$

Step 1.7.6

Simplify the expression.

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Step 1.7.6.1

Move $5$ to the left of $1 + 5 e^{- t}$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 2 e^{- t} (5 \cdot (1 + 5 e^{- t}) \frac{d}{d t} [e^{- t}])}{{(1 + 5 e^{- t})}^{4}}$

Step 1.7.6.2

Multiply $5$ by $- 2$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 10 e^{- t} ((1 + 5 e^{- t}) \frac{d}{d t} [e^{- t}])}{{(1 + 5 e^{- t})}^{4}}$

Step 1.8

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - t$ .

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Step 1.8.1

To apply the Chain Rule, set $u_{3}$ as $- t$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 10 e^{- t} ((1 + 5 e^{- t}) (\frac{d}{d u_{3}} [e^{u_{3}}] \frac{d}{d t} [- t]))}{{(1 + 5 e^{- t})}^{4}}$

Step 1.8.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{3}} [a^{u_{3}}]$ is $a^{u_{3}} \ln (a)$ where $a$ = $e$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 10 e^{- t} ((1 + 5 e^{- t}) (e^{u_{3}} \frac{d}{d t} [- t]))}{{(1 + 5 e^{- t})}^{4}}$

Step 1.8.3

Replace all occurrences of $u_{3}$ with $- t$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 10 e^{- t} ((1 + 5 e^{- t}) (e^{- t} \frac{d}{d t} [- t]))}{{(1 + 5 e^{- t})}^{4}}$

Step 1.9

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 10 e^{- t - t} ((1 + 5 e^{- t}) \frac{d}{d t} [- t])}{{(1 + 5 e^{- t})}^{4}}$

Step 1.10

Subtract $t$ from $- t$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 10 e^{- 2 t} ((1 + 5 e^{- t}) \frac{d}{d t} [- t])}{{(1 + 5 e^{- t})}^{4}}$

Step 1.11

Factor $1 + 5 e^{- t}$ out of $- {(1 + 5 e^{- t})}^{2} e^{- t} - 10 e^{- 2 t} ((1 + 5 e^{- t}) \frac{d}{d t} [- t])$ .

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Step 1.11.1

Factor $1 + 5 e^{- t}$ out of $- {(1 + 5 e^{- t})}^{2} e^{- t}$ .

$25000 \frac{(1 + 5 e^{- t}) (- (1 + 5 e^{- t}) e^{- t}) - 10 e^{- 2 t} ((1 + 5 e^{- t}) \frac{d}{d t} [- t])}{{(1 + 5 e^{- t})}^{4}}$

Step 1.11.2

Factor $1 + 5 e^{- t}$ out of $- 10 e^{- 2 t} ((1 + 5 e^{- t}) \frac{d}{d t} [- t])$ .

$25000 \frac{(1 + 5 e^{- t}) (- (1 + 5 e^{- t}) e^{- t}) + (1 + 5 e^{- t}) (- 10 e^{- 2 t} (\frac{d}{d t} [- t]))}{{(1 + 5 e^{- t})}^{4}}$

Step 1.11.3

Factor $1 + 5 e^{- t}$ out of $(1 + 5 e^{- t}) (- (1 + 5 e^{- t}) e^{- t}) + (1 + 5 e^{- t}) (- 10 e^{- 2 t} (\frac{d}{d t} [- t]))$ .

$25000 \frac{(1 + 5 e^{- t}) (- (1 + 5 e^{- t}) e^{- t} - 10 e^{- 2 t} (\frac{d}{d t} [- t]))}{{(1 + 5 e^{- t})}^{4}}$

Step 1.12

Cancel the common factors.

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Step 1.12.1

Factor $1 + 5 e^{- t}$ out of ${(1 + 5 e^{- t})}^{4}$ .

$25000 \frac{(1 + 5 e^{- t}) (- (1 + 5 e^{- t}) e^{- t} - 10 e^{- 2 t} (\frac{d}{d t} [- t]))}{(1 + 5 e^{- t}) {(1 + 5 e^{- t})}^{3}}$

Step 1.12.2

Cancel the common factor.

$25000 \frac{(1 + 5 e^{- t}) (- (1 + 5 e^{- t}) e^{- t} - 10 e^{- 2 t} (\frac{d}{d t} [- t]))}{(1 + 5 e^{- t}) {(1 + 5 e^{- t})}^{3}}$

Step 1.12.3

Rewrite the expression.

$25000 \frac{- (1 + 5 e^{- t}) e^{- t} - 10 e^{- 2 t} (\frac{d}{d t} [- t])}{{(1 + 5 e^{- t})}^{3}}$

Step 1.13

Since $- 1$ is constant with respect to $t$ , the derivative of $- t$ with respect to $t$ is $- \frac{d}{d t} [t]$ .

$25000 \frac{- (1 + 5 e^{- t}) e^{- t} - 10 e^{- 2 t} (- \frac{d}{d t} [t])}{{(1 + 5 e^{- t})}^{3}}$

Step 1.14

Multiply $- 1$ by $- 10$ .

$25000 \frac{- (1 + 5 e^{- t}) e^{- t} + 10 e^{- 2 t} \frac{d}{d t} [t]}{{(1 + 5 e^{- t})}^{3}}$

Step 1.15

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$25000 \frac{- (1 + 5 e^{- t}) e^{- t} + 10 e^{- 2 t} \cdot 1}{{(1 + 5 e^{- t})}^{3}}$

Step 1.16

Combine fractions.

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Step 1.16.1

Multiply $10$ by $1$ .

$25000 \frac{- (1 + 5 e^{- t}) e^{- t} + 10 e^{- 2 t}}{{(1 + 5 e^{- t})}^{3}}$

Step 1.16.2

Combine $25000$ and $\frac{- (1 + 5 e^{- t}) e^{- t} + 10 e^{- 2 t}}{{(1 + 5 e^{- t})}^{3}}$ .

$\frac{25000 (- (1 + 5 e^{- t}) e^{- t} + 10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 1.17

Simplify.

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Step 1.17.1

Apply the distributive property.

$\frac{25000 ((- 1 \cdot 1 - (5 e^{- t})) e^{- t} + 10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 1.17.2

Apply the distributive property.

$\frac{25000 (- 1 \cdot 1 e^{- t} - (5 e^{- t}) e^{- t} + 10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 1.17.3

Apply the distributive property.

$\frac{25000 (- 1 \cdot 1 e^{- t}) + 25000 (- (5 e^{- t}) e^{- t}) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 1.17.4

Simplify the numerator.

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Step 1.17.4.1

Simplify each term.

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Step 1.17.4.1.1

Multiply $- 1$ by $1$ .

$\frac{25000 (- 1 e^{- t}) + 25000 (- (5 e^{- t}) e^{- t}) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 1.17.4.1.2

Rewrite $- 1 e^{- t}$ as $- e^{- t}$ .

$\frac{25000 (- e^{- t}) + 25000 (- (5 e^{- t}) e^{- t}) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 1.17.4.1.3

Multiply $- 1$ by $25000$ .

$\frac{- 25000 e^{- t} + 25000 (- (5 e^{- t}) e^{- t}) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 1.17.4.1.4

Multiply $e^{- t}$ by $e^{- t}$ by adding the exponents.

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Step 1.17.4.1.4.1

Move $e^{- t}$ .

$\frac{- 25000 e^{- t} + 25000 (- (5 (e^{- t} e^{- t}))) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 1.17.4.1.4.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{- 25000 e^{- t} + 25000 (- (5 e^{- t - t})) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 1.17.4.1.4.3

Subtract $t$ from $- t$ .

$\frac{- 25000 e^{- t} + 25000 (- (5 e^{- 2 t})) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 1.17.4.1.5

Multiply $5$ by $- 1$ .

$\frac{- 25000 e^{- t} + 25000 (- 5 e^{- 2 t}) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 1.17.4.1.6

Multiply $- 5$ by $25000$ .

$\frac{- 25000 e^{- t} - 125000 e^{- 2 t} + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 1.17.4.1.7

Multiply $10$ by $25000$ .

$\frac{- 25000 e^{- t} - 125000 e^{- 2 t} + 250000 e^{- 2 t}}{{(1 + 5 e^{- t})}^{3}}$

Step 1.17.4.2

Add $- 125000 e^{- 2 t}$ and $250000 e^{- 2 t}$ .

$\frac{- 25000 e^{- t} + 125000 e^{- 2 t}}{{(1 + 5 e^{- t})}^{3}}$

Step 1.17.5

Reorder terms.

$\frac{125000 e^{- 2 t} - 25000 e^{- t}}{{(5 e^{- t} + 1)}^{3}}$

Step 1.17.6

Simplify the numerator.

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Step 1.17.6.1

Factor $25000$ out of $125000 e^{- 2 t} - 25000 e^{- t}$ .

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Step 1.17.6.1.1

Factor $25000$ out of $125000 e^{- 2 t}$ .

$\frac{25000 (5 e^{- 2 t}) - 25000 e^{- t}}{{(5 e^{- t} + 1)}^{3}}$

Step 1.17.6.1.2

Factor $25000$ out of $- 25000 e^{- t}$ .

$\frac{25000 (5 e^{- 2 t}) + 25000 (- e^{- t})}{{(5 e^{- t} + 1)}^{3}}$

Step 1.17.6.1.3

Factor $25000$ out of $25000 (5 e^{- 2 t}) + 25000 (- e^{- t})$ .

$\frac{25000 (5 e^{- 2 t} - e^{- t})}{{(5 e^{- t} + 1)}^{3}}$

Step 1.17.6.2

Rewrite $e^{- 2 t}$ as ${(e^{- t})}^{2}$ .

$\frac{25000 (5 {(e^{- t})}^{2} - e^{- t})}{{(5 e^{- t} + 1)}^{3}}$

Step 1.17.6.3

Let $u = e^{- t}$ . Substitute $u$ for all occurrences of $e^{- t}$ .

$\frac{25000 (5 u^{2} - u)}{{(5 e^{- t} + 1)}^{3}}$

Step 1.17.6.4

Factor $u$ out of $5 u^{2} - u$ .

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Step 1.17.6.4.1

Factor $u$ out of $5 u^{2}$ .

$\frac{25000 (u (5 u) - u)}{{(5 e^{- t} + 1)}^{3}}$

Step 1.17.6.4.2

Factor $u$ out of $- u$ .

$\frac{25000 (u (5 u) + u \cdot - 1)}{{(5 e^{- t} + 1)}^{3}}$

Step 1.17.6.4.3

Factor $u$ out of $u (5 u) + u \cdot - 1$ .

$\frac{25000 (u (5 u - 1))}{{(5 e^{- t} + 1)}^{3}}$

Step 1.17.6.5

Replace all occurrences of $u$ with $e^{- t}$ .

$\frac{25000 e^{- t} (5 e^{- t} - 1)}{{(5 e^{- t} + 1)}^{3}}$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $25000$ is constant with respect to $t$ , the derivative of $\frac{25000 e^{- t} (5 e^{- t} - 1)}{{(5 e^{- t} + 1)}^{3}}$ with respect to $t$ is $25000 \frac{d}{d t} [\frac{e^{- t} (5 e^{- t} - 1)}{{(5 e^{- t} + 1)}^{3}}]$ .

$s'' (t) = 25000 \frac{d}{d t} (\frac{e^{- t} (5 e^{- t} - 1)}{{(5 e^{- t} + 1)}^{3}})$

Step 2.2

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} \frac{d}{d t} (e^{- t} (5 e^{- t} - 1)) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{({(5 e^{- t} + 1)}^{3})}^{2}})$

Step 2.3

Multiply the exponents in ${({(5 e^{- t} + 1)}^{3})}^{2}$ .

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Step 2.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} \frac{d}{d t} (e^{- t} (5 e^{- t} - 1)) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{3 \cdot 2}})$

Step 2.3.2

Multiply $3$ by $2$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} \frac{d}{d t} (e^{- t} (5 e^{- t} - 1)) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.4

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = e^{- t}$ and $g (t) = 5 e^{- t} - 1$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (e^{- t} \frac{d}{d t} (5 e^{- t} - 1) + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.5

Differentiate.

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Step 2.5.1

By the Sum Rule, the derivative of $5 e^{- t} - 1$ with respect to $t$ is $\frac{d}{d t} [5 e^{- t}] + \frac{d}{d t} [- 1]$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (e^{- t} (\frac{d}{d t} (5 e^{- t}) + \frac{d}{d t} (- 1)) + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.5.2

Since $5$ is constant with respect to $t$ , the derivative of $5 e^{- t}$ with respect to $t$ is $5 \frac{d}{d t} [e^{- t}]$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (e^{- t} (5 \frac{d}{d t} (e^{- t}) + \frac{d}{d t} (- 1)) + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.6

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - t$ .

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Step 2.6.1

To apply the Chain Rule, set $u_{1}$ as $- t$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (e^{- t} (5 (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d t} (- t)) + \frac{d}{d t} (- 1)) + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.6.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (e^{- t} (5 (e^{u_{1}} \frac{d}{d t} (- t)) + \frac{d}{d t} (- 1)) + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.6.3

Replace all occurrences of $u_{1}$ with $- t$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (e^{- t} (5 (e^{- t} \frac{d}{d t} (- t)) + \frac{d}{d t} (- 1)) + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.7

Differentiate.

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Step 2.7.1

Since $- 1$ is constant with respect to $t$ , the derivative of $- t$ with respect to $t$ is $- \frac{d}{d t} [t]$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (e^{- t} (5 e^{- t} (- \frac{d}{d t} t) + \frac{d}{d t} (- 1)) + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.7.2

Multiply $- 1$ by $5$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (e^{- t} (- 5 e^{- t} \frac{d}{d t} t + \frac{d}{d t} (- 1)) + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.7.3

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (e^{- t} (- 5 e^{- t} \cdot 1 + \frac{d}{d t} (- 1)) + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.7.4

Multiply $- 5$ by $1$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (e^{- t} (- 5 e^{- t} + \frac{d}{d t} (- 1)) + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.7.5

Since $- 1$ is constant with respect to $t$ , the derivative of $- 1$ with respect to $t$ is $0$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (e^{- t} (- 5 e^{- t} + 0) + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.7.6

Add $- 5 e^{- t}$ and $0$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (e^{- t} (- 5 e^{- t}) + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.8

Multiply $e^{- t}$ by $e^{- t}$ by adding the exponents.

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Step 2.8.1

Move $e^{- t}$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (e^{- t} e^{- t} \cdot - 5 + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.8.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (e^{- t - t} \cdot - 5 + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.8.3

Subtract $t$ from $- t$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (e^{- 2 t} \cdot - 5 + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.9

Move $- 5$ to the left of $e^{- 2 t}$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (- 5 \cdot e^{- 2 t} + (5 e^{- t} - 1) \frac{d}{d t} (e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.10

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - t$ .

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Step 2.10.1

To apply the Chain Rule, set $u_{2}$ as $- t$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (- 5 e^{- 2 t} + (5 e^{- t} - 1) (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d t} (- t))) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.10.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (- 5 e^{- 2 t} + (5 e^{- t} - 1) (e^{u_{2}} \frac{d}{d t} (- t))) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.10.3

Replace all occurrences of $u_{2}$ with $- t$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (- 5 e^{- 2 t} + (5 e^{- t} - 1) (e^{- t} \frac{d}{d t} (- t))) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.11

Differentiate.

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Step 2.11.1

Since $- 1$ is constant with respect to $t$ , the derivative of $- t$ with respect to $t$ is $- \frac{d}{d t} [t]$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (- 5 e^{- 2 t} + (5 e^{- t} - 1) e^{- t} (- \frac{d}{d t} t)) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.11.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (- 5 e^{- 2 t} + (5 e^{- t} - 1) e^{- t} (- 1 \cdot 1)) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.11.3

Simplify the expression.

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Step 2.11.3.1

Multiply $- 1$ by $1$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (- 5 e^{- 2 t} + (5 e^{- t} - 1) e^{- t} \cdot - 1) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.11.3.2

Move $- 1$ to the left of $(5 e^{- t} - 1) e^{- t}$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (- 5 e^{- 2 t} - 1 \cdot ((5 e^{- t} - 1) e^{- t})) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.11.3.3

Rewrite $- 1 (5 e^{- t} - 1)$ as $- (5 e^{- t} - 1)$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - e^{- t} (5 e^{- t} - 1) \frac{d}{d t} {(5 e^{- t} + 1)}^{3}}{{(5 e^{- t} + 1)}^{6}})$

Step 2.12

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{3}$ and $g (t) = 5 e^{- t} + 1$ .

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Step 2.12.1

To apply the Chain Rule, set $u_{3}$ as $5 e^{- t} + 1$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - e^{- t} (5 e^{- t} - 1) (\frac{d}{d u (3)} ({u_{3}}^{3}) \frac{d}{d t} (5 e^{- t} + 1))}{{(5 e^{- t} + 1)}^{6}})$

Step 2.12.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{3}} [{u_{3}}^{n}]$ is $n {u_{3}}^{n - 1}$ where $n = 3$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - e^{- t} (5 e^{- t} - 1) (3 {u_{3}}^{2} \frac{d}{d t} (5 e^{- t} + 1))}{{(5 e^{- t} + 1)}^{6}})$

Step 2.12.3

Replace all occurrences of $u_{3}$ with $5 e^{- t} + 1$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - e^{- t} (5 e^{- t} - 1) (3 {(5 e^{- t} + 1)}^{2} \frac{d}{d t} (5 e^{- t} + 1))}{{(5 e^{- t} + 1)}^{6}})$

Step 2.13

Simplify with factoring out.

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Step 2.13.1

Multiply $3$ by $- 1$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{3} (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) ({(5 e^{- t} + 1)}^{2} \frac{d}{d t} (5 e^{- t} + 1))}{{(5 e^{- t} + 1)}^{6}})$

Step 2.13.2

Factor ${(5 e^{- t} + 1)}^{2}$ out of ${(5 e^{- t} + 1)}^{3} (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) ({(5 e^{- t} + 1)}^{2} \frac{d}{d t} [5 e^{- t} + 1])$ .

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Step 2.13.2.1

Factor ${(5 e^{- t} + 1)}^{2}$ out of ${(5 e^{- t} + 1)}^{3} (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t})$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{2} ((5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t})) - 3 e^{- t} (5 e^{- t} - 1) ({(5 e^{- t} + 1)}^{2} \frac{d}{d t} (5 e^{- t} + 1))}{{(5 e^{- t} + 1)}^{6}})$

Step 2.13.2.2

Factor ${(5 e^{- t} + 1)}^{2}$ out of $- 3 e^{- t} (5 e^{- t} - 1) ({(5 e^{- t} + 1)}^{2} \frac{d}{d t} [5 e^{- t} + 1])$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{2} ((5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t})) + {(5 e^{- t} + 1)}^{2} (- 3 e^{- t} (5 e^{- t} - 1) (\frac{d}{d t} (5 e^{- t} + 1)))}{{(5 e^{- t} + 1)}^{6}})$

Step 2.13.2.3

Factor ${(5 e^{- t} + 1)}^{2}$ out of ${(5 e^{- t} + 1)}^{2} ((5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t})) + {(5 e^{- t} + 1)}^{2} (- 3 e^{- t} (5 e^{- t} - 1) (\frac{d}{d t} [5 e^{- t} + 1]))$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{2} ((5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) (\frac{d}{d t} (5 e^{- t} + 1)))}{{(5 e^{- t} + 1)}^{6}})$

Step 2.14

Cancel the common factors.

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Step 2.14.1

Factor ${(5 e^{- t} + 1)}^{2}$ out of ${(5 e^{- t} + 1)}^{6}$ .

$s'' (t) = 25000 (\frac{{(5 e^{- t} + 1)}^{2} ((5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) (\frac{d}{d t} (5 e^{- t} + 1)))}{{(5 e^{- t} + 1)}^{2} {(5 e^{- t} + 1)}^{4}})$

Step 2.14.2

Cancel the common factor.

Step 2.14.3

Rewrite the expression.

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) (\frac{d}{d t} (5 e^{- t} + 1))}{{(5 e^{- t} + 1)}^{4}})$

Step 2.15

Differentiate.

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Step 2.15.1

By the Sum Rule, the derivative of $5 e^{- t} + 1$ with respect to $t$ is $\frac{d}{d t} [5 e^{- t}] + \frac{d}{d t} [1]$ .

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) (\frac{d}{d t} (5 e^{- t}) + \frac{d}{d t} (1))}{{(5 e^{- t} + 1)}^{4}})$

Step 2.15.2

Since $5$ is constant with respect to $t$ , the derivative of $5 e^{- t}$ with respect to $t$ is $5 \frac{d}{d t} [e^{- t}]$ .

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) (5 \frac{d}{d t} (e^{- t}) + \frac{d}{d t} (1))}{{(5 e^{- t} + 1)}^{4}})$

Step 2.16

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - t$ .

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Step 2.16.1

To apply the Chain Rule, set $u_{4}$ as $- t$ .

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) (5 (\frac{d}{d u (4)} (e^{u_{4}}) \frac{d}{d t} (- t)) + \frac{d}{d t} (1))}{{(5 e^{- t} + 1)}^{4}})$

Step 2.16.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{4}} [a^{u_{4}}]$ is $a^{u_{4}} \ln (a)$ where $a$ = $e$ .

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) (5 (e^{u_{4}} \frac{d}{d t} (- t)) + \frac{d}{d t} (1))}{{(5 e^{- t} + 1)}^{4}})$

Step 2.16.3

Replace all occurrences of $u_{4}$ with $- t$ .

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) (5 (e^{- t} \frac{d}{d t} (- t)) + \frac{d}{d t} (1))}{{(5 e^{- t} + 1)}^{4}})$

Step 2.17

Differentiate.

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Step 2.17.1

Since $- 1$ is constant with respect to $t$ , the derivative of $- t$ with respect to $t$ is $- \frac{d}{d t} [t]$ .

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) (5 e^{- t} (- \frac{d}{d t} t) + \frac{d}{d t} (1))}{{(5 e^{- t} + 1)}^{4}})$

Step 2.17.2

Multiply $- 1$ by $5$ .

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) (- 5 e^{- t} \frac{d}{d t} t + \frac{d}{d t} (1))}{{(5 e^{- t} + 1)}^{4}})$

Step 2.17.3

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) (- 5 e^{- t} \cdot 1 + \frac{d}{d t} (1))}{{(5 e^{- t} + 1)}^{4}})$

Step 2.17.4

Multiply $- 5$ by $1$ .

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) (- 5 e^{- t} + \frac{d}{d t} (1))}{{(5 e^{- t} + 1)}^{4}})$

Step 2.17.5

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) (- 5 e^{- t} + 0)}{{(5 e^{- t} + 1)}^{4}})$

Step 2.17.6

Simplify the expression.

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Step 2.17.6.1

Add $- 5 e^{- t}$ and $0$ .

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) - 3 e^{- t} (5 e^{- t} - 1) (- 5 e^{- t})}{{(5 e^{- t} + 1)}^{4}})$

Step 2.17.6.2

Multiply $- 5$ by $- 3$ .

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) + 15 e^{- t} (5 e^{- t} - 1) e^{- t}}{{(5 e^{- t} + 1)}^{4}})$

Step 2.18

Multiply $e^{- t}$ by $e^{- t}$ by adding the exponents.

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Step 2.18.1

Move $e^{- t}$ .

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) + 15 (e^{- t} e^{- t}) (5 e^{- t} - 1)}{{(5 e^{- t} + 1)}^{4}})$

Step 2.18.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) + 15 e^{- t - t} (5 e^{- t} - 1)}{{(5 e^{- t} + 1)}^{4}})$

Step 2.18.3

Subtract $t$ from $- t$ .

$s'' (t) = 25000 (\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) + 15 e^{- 2 t} (5 e^{- t} - 1)}{{(5 e^{- t} + 1)}^{4}})$

Step 2.19

Combine $25000$ and $\frac{(5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) + 15 e^{- 2 t} (5 e^{- t} - 1)}{{(5 e^{- t} + 1)}^{4}}$ .

$s'' (t) = \frac{25000 ((5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t} - 1) e^{- t}) + 15 e^{- 2 t} (5 e^{- t} - 1))}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20

Simplify.

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Step 2.20.1

Apply the distributive property.

$s'' (t) = \frac{25000 ((5 e^{- t} + 1) (- 5 e^{- 2 t} + (- (5 e^{- t}) + 1) e^{- t}) + 15 e^{- 2 t} (5 e^{- t} - 1))}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.2

Apply the distributive property.

$s'' (t) = \frac{25000 ((5 e^{- t} + 1) (- 5 e^{- 2 t} - 5 e^{- t} e^{- t} + 1 e^{- t}) + 15 e^{- 2 t} (5 e^{- t} - 1))}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.3

Apply the distributive property.

$s'' (t) = \frac{25000 ((5 e^{- t} + 1) (- 5 e^{- 2 t} - 5 e^{- t} e^{- t} + 1 e^{- t}) + 15 e^{- 2 t} (5 e^{- t}) + 15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.4

Apply the distributive property.

$s'' (t) = \frac{25000 ((5 e^{- t} + 1) (- 5 e^{- 2 t} - 5 e^{- t} e^{- t} + 1 e^{- t})) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5

Simplify the numerator.

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Step 2.20.5.1

Simplify each term.

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Step 2.20.5.1.1

Simplify each term.

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Step 2.20.5.1.1.1

Multiply $e^{- t}$ by $e^{- t}$ by adding the exponents.

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Step 2.20.5.1.1.1.1

Move $e^{- t}$ .

$s'' (t) = \frac{25000 ((5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 (e^{- t} e^{- t})) + 1 e^{- t})) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.1.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$s'' (t) = \frac{25000 ((5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- t - t}) + 1 e^{- t})) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.1.1.3

Subtract $t$ from $- t$ .

$s'' (t) = \frac{25000 ((5 e^{- t} + 1) (- 5 e^{- 2 t} - (5 e^{- 2 t}) + 1 e^{- t})) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.1.2

Multiply $5$ by $- 1$ .

$s'' (t) = \frac{25000 ((5 e^{- t} + 1) (- 5 e^{- 2 t} - 5 e^{- 2 t} + 1 e^{- t})) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.1.3

Multiply $- 1$ by $- 1$ .

$s'' (t) = \frac{25000 ((5 e^{- t} + 1) (- 5 e^{- 2 t} - 5 e^{- 2 t} + 1 e^{- t})) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.1.4

Multiply $e^{- t}$ by $1$ .

$s'' (t) = \frac{25000 ((5 e^{- t} + 1) (- 5 e^{- 2 t} - 5 e^{- 2 t} + e^{- t})) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.2

Subtract $5 e^{- 2 t}$ from $- 5 e^{- 2 t}$ .

$s'' (t) = \frac{25000 ((5 e^{- t} + 1) (- 10 e^{- 2 t} + e^{- t})) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.3

Expand $(5 e^{- t} + 1) (- 10 e^{- 2 t} + e^{- t})$ using the FOIL Method.

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Step 2.20.5.1.3.1

Apply the distributive property.

$s'' (t) = \frac{25000 (5 e^{- t} (- 10 e^{- 2 t} + e^{- t}) + 1 (- 10 e^{- 2 t} + e^{- t})) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.3.2

Apply the distributive property.

$s'' (t) = \frac{25000 (5 e^{- t} (- 10 e^{- 2 t}) + 5 e^{- t} e^{- t} + 1 (- 10 e^{- 2 t} + e^{- t})) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.3.3

Apply the distributive property.

$s'' (t) = \frac{25000 (5 e^{- t} (- 10 e^{- 2 t}) + 5 e^{- t} e^{- t} + 1 (- 10 e^{- 2 t}) + 1 e^{- t}) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.4

Simplify and combine like terms.

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Step 2.20.5.1.4.1

Simplify each term.

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Step 2.20.5.1.4.1.1

Rewrite using the commutative property of multiplication.

$s'' (t) = \frac{25000 (5 \cdot (- 10 e^{- t} e^{- 2 t}) + 5 e^{- t} e^{- t} + 1 (- 10 e^{- 2 t}) + 1 e^{- t}) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.4.1.2

Multiply $e^{- t}$ by $e^{- 2 t}$ by adding the exponents.

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Step 2.20.5.1.4.1.2.1

Move $e^{- 2 t}$ .

$s'' (t) = \frac{25000 (5 \cdot (- 10 (e^{- 2 t} e^{- t})) + 5 e^{- t} e^{- t} + 1 (- 10 e^{- 2 t}) + 1 e^{- t}) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.4.1.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$s'' (t) = \frac{25000 (5 \cdot (- 10 e^{- 2 t - t}) + 5 e^{- t} e^{- t} + 1 (- 10 e^{- 2 t}) + 1 e^{- t}) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.4.1.2.3

Subtract $t$ from $- 2 t$ .

$s'' (t) = \frac{25000 (5 \cdot (- 10 e^{- 3 t}) + 5 e^{- t} e^{- t} + 1 (- 10 e^{- 2 t}) + 1 e^{- t}) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.4.1.3

Multiply $5$ by $- 10$ .

$s'' (t) = \frac{25000 (- 50 e^{- 3 t} + 5 e^{- t} e^{- t} + 1 (- 10 e^{- 2 t}) + 1 e^{- t}) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.4.1.4

Multiply $e^{- t}$ by $e^{- t}$ by adding the exponents.

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Step 2.20.5.1.4.1.4.1

Move $e^{- t}$ .

$s'' (t) = \frac{25000 (- 50 e^{- 3 t} + 5 (e^{- t} e^{- t}) + 1 (- 10 e^{- 2 t}) + 1 e^{- t}) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.4.1.4.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$s'' (t) = \frac{25000 (- 50 e^{- 3 t} + 5 e^{- t - t} + 1 (- 10 e^{- 2 t}) + 1 e^{- t}) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.4.1.4.3

Subtract $t$ from $- t$ .

$s'' (t) = \frac{25000 (- 50 e^{- 3 t} + 5 e^{- 2 t} + 1 (- 10 e^{- 2 t}) + 1 e^{- t}) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.4.1.5

Multiply $- 10 e^{- 2 t}$ by $1$ .

$s'' (t) = \frac{25000 (- 50 e^{- 3 t} + 5 e^{- 2 t} - 10 e^{- 2 t} + 1 e^{- t}) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.4.1.6

Multiply $e^{- t}$ by $1$ .

$s'' (t) = \frac{25000 (- 50 e^{- 3 t} + 5 e^{- 2 t} - 10 e^{- 2 t} + e^{- t}) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.4.2

Subtract $10 e^{- 2 t}$ from $5 e^{- 2 t}$ .

$s'' (t) = \frac{25000 (- 50 e^{- 3 t} - 5 e^{- 2 t} + e^{- t}) + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.5

Apply the distributive property.

$s'' (t) = \frac{25000 (- 50 e^{- 3 t}) + 25000 (- 5 e^{- 2 t}) + 25000 e^{- t} + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.6

Simplify.

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Step 2.20.5.1.6.1

Multiply $- 50$ by $25000$ .

$s'' (t) = \frac{- 1250000 e^{- 3 t} + 25000 (- 5 e^{- 2 t}) + 25000 e^{- t} + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.6.2

Multiply $- 5$ by $25000$ .

$s'' (t) = \frac{- 1250000 e^{- 3 t} - 125000 e^{- 2 t} + 25000 e^{- t} + 25000 (15 e^{- 2 t} (5 e^{- t})) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.7

Multiply $e^{- 2 t}$ by $e^{- t}$ by adding the exponents.

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Step 2.20.5.1.7.1

Move $e^{- t}$ .

$s'' (t) = \frac{- 1250000 e^{- 3 t} - 125000 e^{- 2 t} + 25000 e^{- t} + 25000 (15 (e^{- t} e^{- 2 t}) \cdot 5) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.7.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$s'' (t) = \frac{- 1250000 e^{- 3 t} - 125000 e^{- 2 t} + 25000 e^{- t} + 25000 (15 e^{- t - 2 t} \cdot 5) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.7.3

Subtract $2 t$ from $- t$ .

$s'' (t) = \frac{- 1250000 e^{- 3 t} - 125000 e^{- 2 t} + 25000 e^{- t} + 25000 (15 e^{- 3 t} \cdot 5) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.8

Multiply $5$ by $15$ .

$s'' (t) = \frac{- 1250000 e^{- 3 t} - 125000 e^{- 2 t} + 25000 e^{- t} + 25000 (75 e^{- 3 t}) + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.9

Multiply $75$ by $25000$ .

$s'' (t) = \frac{- 1250000 e^{- 3 t} - 125000 e^{- 2 t} + 25000 e^{- t} + 1875000 e^{- 3 t} + 25000 (15 e^{- 2 t} \cdot - 1)}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.10

Multiply $- 1$ by $15$ .

$s'' (t) = \frac{- 1250000 e^{- 3 t} - 125000 e^{- 2 t} + 25000 e^{- t} + 1875000 e^{- 3 t} + 25000 (- 15 e^{- 2 t})}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.1.11

Multiply $- 15$ by $25000$ .

$s'' (t) = \frac{- 1250000 e^{- 3 t} - 125000 e^{- 2 t} + 25000 e^{- t} + 1875000 e^{- 3 t} - 375000 e^{- 2 t}}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.2

Add $- 1250000 e^{- 3 t}$ and $1875000 e^{- 3 t}$ .

$s'' (t) = \frac{625000 e^{- 3 t} - 125000 e^{- 2 t} + 25000 e^{- t} - 375000 e^{- 2 t}}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.5.3

Subtract $375000 e^{- 2 t}$ from $- 125000 e^{- 2 t}$ .

$s'' (t) = \frac{625000 e^{- 3 t} + 25000 e^{- t} - 500000 e^{- 2 t}}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.6

Factor $25000$ out of $625000 e^{- 3 t} + 25000 e^{- t} - 500000 e^{- 2 t}$ .

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Step 2.20.6.1

Factor $25000$ out of $625000 e^{- 3 t}$ .

$s'' (t) = \frac{25000 (25 e^{- 3 t}) + 25000 e^{- t} - 500000 e^{- 2 t}}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.6.2

Factor $25000$ out of $25000 e^{- t}$ .

$s'' (t) = \frac{25000 (25 e^{- 3 t}) + 25000 (e^{- t}) - 500000 e^{- 2 t}}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.6.3

Factor $25000$ out of $- 500000 e^{- 2 t}$ .

$s'' (t) = \frac{25000 (25 e^{- 3 t}) + 25000 (e^{- t}) + 25000 (- 20 e^{- 2 t})}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.6.4

Factor $25000$ out of $25000 (25 e^{- 3 t}) + 25000 (e^{- t})$ .

$s'' (t) = \frac{25000 (25 e^{- 3 t} + e^{- t}) + 25000 (- 20 e^{- 2 t})}{{(5 e^{- t} + 1)}^{4}}$

Step 2.20.6.5

Factor $25000$ out of $25000 (25 e^{- 3 t} + e^{- t}) + 25000 (- 20 e^{- 2 t})$ .

$s'' (t) = \frac{25000 (25 e^{- 3 t} + e^{- t} - 20 e^{- 2 t})}{{(5 e^{- t} + 1)}^{4}}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$\frac{25000 e^{- t} (5 e^{- t} - 1)}{{(5 e^{- t} + 1)}^{3}} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

Since $25000$ is constant with respect to $t$ , the derivative of $\frac{25000 e^{- t}}{{(1 + 5 e^{- t})}^{2}}$ with respect to $t$ is $25000 \frac{d}{d t} [\frac{e^{- t}}{{(1 + 5 e^{- t})}^{2}}]$ .

$25000 \frac{d}{d t} [\frac{e^{- t}}{{(1 + 5 e^{- t})}^{2}}]$

Step 4.1.2

$25000 \frac{{(1 + 5 e^{- t})}^{2} \frac{d}{d t} [e^{- t}] - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{({(1 + 5 e^{- t})}^{2})}^{2}}$

Step 4.1.3

Multiply the exponents in ${({(1 + 5 e^{- t})}^{2})}^{2}$ .

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Step 4.1.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} \frac{d}{d t} [e^{- t}] - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{2 \cdot 2}}$

Step 4.1.3.2

Multiply $2$ by $2$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} \frac{d}{d t} [e^{- t}] - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.4

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - t$ .

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Step 4.1.4.1

To apply the Chain Rule, set $u_{1}$ as $- t$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d t} [- t]) - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.4.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} (e^{u_{1}} \frac{d}{d t} [- t]) - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.4.3

Replace all occurrences of $u_{1}$ with $- t$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} (e^{- t} \frac{d}{d t} [- t]) - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.5

Differentiate.

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Step 4.1.5.1

Since $- 1$ is constant with respect to $t$ , the derivative of $- t$ with respect to $t$ is $- \frac{d}{d t} [t]$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} e^{- t} (- \frac{d}{d t} [t]) - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.5.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} e^{- t} (- 1 \cdot 1) - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.5.3

Simplify the expression.

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Step 4.1.5.3.1

Multiply $- 1$ by $1$ .

$25000 \frac{{(1 + 5 e^{- t})}^{2} e^{- t} \cdot - 1 - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.5.3.2

Move $- 1$ to the left of ${(1 + 5 e^{- t})}^{2} e^{- t}$ .

$25000 \frac{- 1 \cdot ({(1 + 5 e^{- t})}^{2} e^{- t}) - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.5.3.3

Rewrite $- 1 {(1 + 5 e^{- t})}^{2}$ as $- {(1 + 5 e^{- t})}^{2}$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - e^{- t} \frac{d}{d t} [{(1 + 5 e^{- t})}^{2}]}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.6

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{2}$ and $g (t) = 1 + 5 e^{- t}$ .

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Step 4.1.6.1

To apply the Chain Rule, set $u_{2}$ as $1 + 5 e^{- t}$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - e^{- t} (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d t} [1 + 5 e^{- t}])}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.6.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 2$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - e^{- t} (2 u_{2} \frac{d}{d t} [1 + 5 e^{- t}])}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.6.3

Replace all occurrences of $u_{2}$ with $1 + 5 e^{- t}$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - e^{- t} (2 (1 + 5 e^{- t}) \frac{d}{d t} [1 + 5 e^{- t}])}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.7

Differentiate.

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Step 4.1.7.1

Multiply $2$ by $- 1$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 2 e^{- t} ((1 + 5 e^{- t}) \frac{d}{d t} [1 + 5 e^{- t}])}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.7.2

By the Sum Rule, the derivative of $1 + 5 e^{- t}$ with respect to $t$ is $\frac{d}{d t} [1] + \frac{d}{d t} [5 e^{- t}]$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 2 e^{- t} ((1 + 5 e^{- t}) (\frac{d}{d t} [1] + \frac{d}{d t} [5 e^{- t}]))}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.7.3

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 2 e^{- t} ((1 + 5 e^{- t}) (0 + \frac{d}{d t} [5 e^{- t}]))}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.7.4

Add $0$ and $\frac{d}{d t} [5 e^{- t}]$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 2 e^{- t} ((1 + 5 e^{- t}) \frac{d}{d t} [5 e^{- t}])}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.7.5

Since $5$ is constant with respect to $t$ , the derivative of $5 e^{- t}$ with respect to $t$ is $5 \frac{d}{d t} [e^{- t}]$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 2 e^{- t} ((1 + 5 e^{- t}) (5 \frac{d}{d t} [e^{- t}]))}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.7.6

Simplify the expression.

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Step 4.1.7.6.1

Move $5$ to the left of $1 + 5 e^{- t}$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 2 e^{- t} (5 \cdot (1 + 5 e^{- t}) \frac{d}{d t} [e^{- t}])}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.7.6.2

Multiply $5$ by $- 2$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 10 e^{- t} ((1 + 5 e^{- t}) \frac{d}{d t} [e^{- t}])}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.8

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - t$ .

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Step 4.1.8.1

To apply the Chain Rule, set $u_{3}$ as $- t$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 10 e^{- t} ((1 + 5 e^{- t}) (\frac{d}{d u_{3}} [e^{u_{3}}] \frac{d}{d t} [- t]))}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.8.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{3}} [a^{u_{3}}]$ is $a^{u_{3}} \ln (a)$ where $a$ = $e$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 10 e^{- t} ((1 + 5 e^{- t}) (e^{u_{3}} \frac{d}{d t} [- t]))}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.8.3

Replace all occurrences of $u_{3}$ with $- t$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 10 e^{- t} ((1 + 5 e^{- t}) (e^{- t} \frac{d}{d t} [- t]))}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.9

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 10 e^{- t - t} ((1 + 5 e^{- t}) \frac{d}{d t} [- t])}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.10

Subtract $t$ from $- t$ .

$25000 \frac{- {(1 + 5 e^{- t})}^{2} e^{- t} - 10 e^{- 2 t} ((1 + 5 e^{- t}) \frac{d}{d t} [- t])}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.11

Factor $1 + 5 e^{- t}$ out of $- {(1 + 5 e^{- t})}^{2} e^{- t} - 10 e^{- 2 t} ((1 + 5 e^{- t}) \frac{d}{d t} [- t])$ .

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Step 4.1.11.1

Factor $1 + 5 e^{- t}$ out of $- {(1 + 5 e^{- t})}^{2} e^{- t}$ .

$25000 \frac{(1 + 5 e^{- t}) (- (1 + 5 e^{- t}) e^{- t}) - 10 e^{- 2 t} ((1 + 5 e^{- t}) \frac{d}{d t} [- t])}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.11.2

Factor $1 + 5 e^{- t}$ out of $- 10 e^{- 2 t} ((1 + 5 e^{- t}) \frac{d}{d t} [- t])$ .

$25000 \frac{(1 + 5 e^{- t}) (- (1 + 5 e^{- t}) e^{- t}) + (1 + 5 e^{- t}) (- 10 e^{- 2 t} (\frac{d}{d t} [- t]))}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.11.3

Factor $1 + 5 e^{- t}$ out of $(1 + 5 e^{- t}) (- (1 + 5 e^{- t}) e^{- t}) + (1 + 5 e^{- t}) (- 10 e^{- 2 t} (\frac{d}{d t} [- t]))$ .

$25000 \frac{(1 + 5 e^{- t}) (- (1 + 5 e^{- t}) e^{- t} - 10 e^{- 2 t} (\frac{d}{d t} [- t]))}{{(1 + 5 e^{- t})}^{4}}$

Step 4.1.12

Cancel the common factors.

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Step 4.1.12.1

Factor $1 + 5 e^{- t}$ out of ${(1 + 5 e^{- t})}^{4}$ .

$25000 \frac{(1 + 5 e^{- t}) (- (1 + 5 e^{- t}) e^{- t} - 10 e^{- 2 t} (\frac{d}{d t} [- t]))}{(1 + 5 e^{- t}) {(1 + 5 e^{- t})}^{3}}$

Step 4.1.12.2

Cancel the common factor.

$25000 \frac{(1 + 5 e^{- t}) (- (1 + 5 e^{- t}) e^{- t} - 10 e^{- 2 t} (\frac{d}{d t} [- t]))}{(1 + 5 e^{- t}) {(1 + 5 e^{- t})}^{3}}$

Step 4.1.12.3

Rewrite the expression.

$25000 \frac{- (1 + 5 e^{- t}) e^{- t} - 10 e^{- 2 t} (\frac{d}{d t} [- t])}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.13

Since $- 1$ is constant with respect to $t$ , the derivative of $- t$ with respect to $t$ is $- \frac{d}{d t} [t]$ .

$25000 \frac{- (1 + 5 e^{- t}) e^{- t} - 10 e^{- 2 t} (- \frac{d}{d t} [t])}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.14

Multiply $- 1$ by $- 10$ .

$25000 \frac{- (1 + 5 e^{- t}) e^{- t} + 10 e^{- 2 t} \frac{d}{d t} [t]}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.15

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$25000 \frac{- (1 + 5 e^{- t}) e^{- t} + 10 e^{- 2 t} \cdot 1}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.16

Combine fractions.

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Step 4.1.16.1

Multiply $10$ by $1$ .

$25000 \frac{- (1 + 5 e^{- t}) e^{- t} + 10 e^{- 2 t}}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.16.2

Combine $25000$ and $\frac{- (1 + 5 e^{- t}) e^{- t} + 10 e^{- 2 t}}{{(1 + 5 e^{- t})}^{3}}$ .

$\frac{25000 (- (1 + 5 e^{- t}) e^{- t} + 10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.17

Simplify.

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Step 4.1.17.1

Apply the distributive property.

$\frac{25000 ((- 1 \cdot 1 - (5 e^{- t})) e^{- t} + 10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.17.2

Apply the distributive property.

$\frac{25000 (- 1 \cdot 1 e^{- t} - (5 e^{- t}) e^{- t} + 10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.17.3

Apply the distributive property.

$\frac{25000 (- 1 \cdot 1 e^{- t}) + 25000 (- (5 e^{- t}) e^{- t}) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.17.4

Simplify the numerator.

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Step 4.1.17.4.1

Simplify each term.

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Step 4.1.17.4.1.1

Multiply $- 1$ by $1$ .

$\frac{25000 (- 1 e^{- t}) + 25000 (- (5 e^{- t}) e^{- t}) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.17.4.1.2

Rewrite $- 1 e^{- t}$ as $- e^{- t}$ .

$\frac{25000 (- e^{- t}) + 25000 (- (5 e^{- t}) e^{- t}) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.17.4.1.3

Multiply $- 1$ by $25000$ .

$\frac{- 25000 e^{- t} + 25000 (- (5 e^{- t}) e^{- t}) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.17.4.1.4

Multiply $e^{- t}$ by $e^{- t}$ by adding the exponents.

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Step 4.1.17.4.1.4.1

Move $e^{- t}$ .

$\frac{- 25000 e^{- t} + 25000 (- (5 (e^{- t} e^{- t}))) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.17.4.1.4.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{- 25000 e^{- t} + 25000 (- (5 e^{- t - t})) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.17.4.1.4.3

Subtract $t$ from $- t$ .

$\frac{- 25000 e^{- t} + 25000 (- (5 e^{- 2 t})) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.17.4.1.5

Multiply $5$ by $- 1$ .

$\frac{- 25000 e^{- t} + 25000 (- 5 e^{- 2 t}) + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.17.4.1.6

Multiply $- 5$ by $25000$ .

$\frac{- 25000 e^{- t} - 125000 e^{- 2 t} + 25000 (10 e^{- 2 t})}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.17.4.1.7

Multiply $10$ by $25000$ .

$\frac{- 25000 e^{- t} - 125000 e^{- 2 t} + 250000 e^{- 2 t}}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.17.4.2

Add $- 125000 e^{- 2 t}$ and $250000 e^{- 2 t}$ .

$\frac{- 25000 e^{- t} + 125000 e^{- 2 t}}{{(1 + 5 e^{- t})}^{3}}$

Step 4.1.17.5

Reorder terms.

$\frac{125000 e^{- 2 t} - 25000 e^{- t}}{{(5 e^{- t} + 1)}^{3}}$

Step 4.1.17.6

Simplify the numerator.

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Step 4.1.17.6.1

Factor $25000$ out of $125000 e^{- 2 t} - 25000 e^{- t}$ .

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Step 4.1.17.6.1.1

Factor $25000$ out of $125000 e^{- 2 t}$ .

$\frac{25000 (5 e^{- 2 t}) - 25000 e^{- t}}{{(5 e^{- t} + 1)}^{3}}$

Step 4.1.17.6.1.2

Factor $25000$ out of $- 25000 e^{- t}$ .

$\frac{25000 (5 e^{- 2 t}) + 25000 (- e^{- t})}{{(5 e^{- t} + 1)}^{3}}$

Step 4.1.17.6.1.3

Factor $25000$ out of $25000 (5 e^{- 2 t}) + 25000 (- e^{- t})$ .

$\frac{25000 (5 e^{- 2 t} - e^{- t})}{{(5 e^{- t} + 1)}^{3}}$

Step 4.1.17.6.2

Rewrite $e^{- 2 t}$ as ${(e^{- t})}^{2}$ .

$\frac{25000 (5 {(e^{- t})}^{2} - e^{- t})}{{(5 e^{- t} + 1)}^{3}}$

Step 4.1.17.6.3

Let $u = e^{- t}$ . Substitute $u$ for all occurrences of $e^{- t}$ .

$\frac{25000 (5 u^{2} - u)}{{(5 e^{- t} + 1)}^{3}}$

Step 4.1.17.6.4

Factor $u$ out of $5 u^{2} - u$ .

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Step 4.1.17.6.4.1

Factor $u$ out of $5 u^{2}$ .

$\frac{25000 (u (5 u) - u)}{{(5 e^{- t} + 1)}^{3}}$

Step 4.1.17.6.4.2

Factor $u$ out of $- u$ .

$\frac{25000 (u (5 u) + u \cdot - 1)}{{(5 e^{- t} + 1)}^{3}}$

Step 4.1.17.6.4.3

Factor $u$ out of $u (5 u) + u \cdot - 1$ .

$\frac{25000 (u (5 u - 1))}{{(5 e^{- t} + 1)}^{3}}$

Step 4.1.17.6.5

Replace all occurrences of $u$ with $e^{- t}$ .

$f' (t) = \frac{25000 e^{- t} (5 e^{- t} - 1)}{{(5 e^{- t} + 1)}^{3}}$

Step 4.2

The first derivative of $s (t)$ with respect to $t$ is $\frac{25000 e^{- t} (5 e^{- t} - 1)}{{(5 e^{- t} + 1)}^{3}}$ .

$\frac{25000 e^{- t} (5 e^{- t} - 1)}{{(5 e^{- t} + 1)}^{3}}$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{25000 e^{- t} (5 e^{- t} - 1)}{{(5 e^{- t} + 1)}^{3}} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$\frac{25000 e^{- t} (5 e^{- t} - 1)}{{(5 e^{- t} + 1)}^{3}} = 0$

Step 5.2

Set the numerator equal to zero.

$25000 e^{- t} (5 e^{- t} - 1) = 0$

Step 5.3

Solve the equation for $t$ .

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Step 5.3.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$e^{- t} = 0$

$5 e^{- t} - 1 = 0$

Step 5.3.2

Set $e^{- t}$ equal to $0$ and solve for $t$ .

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Step 5.3.2.1

Set $e^{- t}$ equal to $0$ .

$e^{- t} = 0$

Step 5.3.2.2

Solve $e^{- t} = 0$ for $t$ .

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Step 5.3.2.2.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{- t}) = \ln (0)$

Step 5.3.2.2.2

The equation cannot be solved because $\ln (0)$ is undefined.

Undefined

Step 5.3.2.2.3

There is no solution for $e^{- t} = 0$

No solution

Step 5.3.3

Set $5 e^{- t} - 1$ equal to $0$ and solve for $t$ .

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Step 5.3.3.1

Set $5 e^{- t} - 1$ equal to $0$ .

$5 e^{- t} - 1 = 0$

Step 5.3.3.2

Solve $5 e^{- t} - 1 = 0$ for $t$ .

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Step 5.3.3.2.1

Add $1$ to both sides of the equation.

$5 e^{- t} = 1$

Step 5.3.3.2.2

Divide each term in $5 e^{- t} = 1$ by $5$ and simplify.

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Step 5.3.3.2.2.1

Divide each term in $5 e^{- t} = 1$ by $5$ .

$\frac{5 e^{- t}}{5} = \frac{1}{5}$

Step 5.3.3.2.2.2

Simplify the left side.

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Step 5.3.3.2.2.2.1

Cancel the common factor of $5$ .

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Step 5.3.3.2.2.2.1.1

Cancel the common factor.

$\frac{5 e^{- t}}{5} = \frac{1}{5}$

Step 5.3.3.2.2.2.1.2

Divide $e^{- t}$ by $1$ .

$e^{- t} = \frac{1}{5}$

Step 5.3.3.2.3

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{- t}) = \ln (\frac{1}{5})$

Step 5.3.3.2.4

Expand the left side.

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Step 5.3.3.2.4.1

Expand $\ln (e^{- t})$ by moving $- t$ outside the logarithm.

$- t \ln (e) = \ln (\frac{1}{5})$

Step 5.3.3.2.4.2

The natural logarithm of $e$ is $1$ .

$- t \cdot 1 = \ln (\frac{1}{5})$

Step 5.3.3.2.4.3

Multiply $- 1$ by $1$ .

$- t = \ln (\frac{1}{5})$

Step 5.3.3.2.5

Divide each term in $- t = \ln (\frac{1}{5})$ by $- 1$ and simplify.

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Step 5.3.3.2.5.1

Divide each term in $- t = \ln (\frac{1}{5})$ by $- 1$ .

$\frac{- t}{- 1} = \frac{\ln (\frac{1}{5})}{- 1}$

Step 5.3.3.2.5.2

Simplify the left side.

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Step 5.3.3.2.5.2.1

Dividing two negative values results in a positive value.

$\frac{t}{1} = \frac{\ln (\frac{1}{5})}{- 1}$

Step 5.3.3.2.5.2.2

Divide $t$ by $1$ .

$t = \frac{\ln (\frac{1}{5})}{- 1}$

Step 5.3.3.2.5.3

Simplify the right side.

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Step 5.3.3.2.5.3.1

Move the negative one from the denominator of $\frac{\ln (\frac{1}{5})}{- 1}$ .

$t = - 1 \cdot \ln (\frac{1}{5})$

Step 5.3.3.2.5.3.2

Rewrite $- 1 \cdot \ln (\frac{1}{5})$ as $- \ln (\frac{1}{5})$ .

$t = - \ln (\frac{1}{5})$

Step 5.3.4

The final solution is all the values that make $25000 e^{- t} (5 e^{- t} - 1) = 0$ true.

$t = - \ln (\frac{1}{5})$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$t = - \ln (\frac{1}{5})$

Step 8

Evaluate the second derivative at $t = - \ln (\frac{1}{5})$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{25000 (25 e^{- 3 (- \ln (\frac{1}{5}))} + e^{- (- \ln (\frac{1}{5}))} - 20 e^{- 2 (- \ln (\frac{1}{5}))})}{{(5 e^{- (- \ln (\frac{1}{5}))} + 1)}^{4}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the numerator.

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Step 9.1.1

Multiply $- 3 (- \ln (\frac{1}{5}))$ .

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Step 9.1.1.1

Multiply $- 1$ by $- 3$ .

$\frac{25000 (25 e^{3 \ln (\frac{1}{5})} + e^{- - \ln (\frac{1}{5})} - 20 e^{- 2 (- \ln (\frac{1}{5}))})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.1.2

Simplify $3 \ln (\frac{1}{5})$ by moving $3$ inside the logarithm.

$\frac{25000 (25 e^{\ln ({(\frac{1}{5})}^{3})} + e^{- - \ln (\frac{1}{5})} - 20 e^{- 2 (- \ln (\frac{1}{5}))})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.2

Exponentiation and log are inverse functions.

$\frac{25000 (25 {(\frac{1}{5})}^{3} + e^{- - \ln (\frac{1}{5})} - 20 e^{- 2 (- \ln (\frac{1}{5}))})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.3

Apply the product rule to $\frac{1}{5}$ .

$\frac{25000 (25 \frac{1^{3}}{5^{3}} + e^{- - \ln (\frac{1}{5})} - 20 e^{- 2 (- \ln (\frac{1}{5}))})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.4

One to any power is one.

$\frac{25000 (25 \frac{1}{5^{3}} + e^{- - \ln (\frac{1}{5})} - 20 e^{- 2 (- \ln (\frac{1}{5}))})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.5

Raise $5$ to the power of $3$ .

$\frac{25000 (25 (\frac{1}{125}) + e^{- - \ln (\frac{1}{5})} - 20 e^{- 2 (- \ln (\frac{1}{5}))})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.6

Cancel the common factor of $25$ .

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Step 9.1.6.1

Factor $25$ out of $125$ .

$\frac{25000 (25 \frac{1}{25 (5)} + e^{- - \ln (\frac{1}{5})} - 20 e^{- 2 (- \ln (\frac{1}{5}))})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.6.2

Cancel the common factor.

$\frac{25000 (25 \frac{1}{25 \cdot 5} + e^{- - \ln (\frac{1}{5})} - 20 e^{- 2 (- \ln (\frac{1}{5}))})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.6.3

Rewrite the expression.

$\frac{25000 (\frac{1}{5} + e^{- - \ln (\frac{1}{5})} - 20 e^{- 2 (- \ln (\frac{1}{5}))})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.7

Multiply $- - \ln (\frac{1}{5})$ .

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Step 9.1.7.1

Multiply $- 1$ by $- 1$ .

$\frac{25000 (\frac{1}{5} + e^{1 \ln (\frac{1}{5})} - 20 e^{- 2 (- \ln (\frac{1}{5}))})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.7.2

Multiply $\ln (\frac{1}{5})$ by $1$ .

$\frac{25000 (\frac{1}{5} + e^{\ln (\frac{1}{5})} - 20 e^{- 2 (- \ln (\frac{1}{5}))})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.8

Exponentiation and log are inverse functions.

$\frac{25000 (\frac{1}{5} + \frac{1}{5} - 20 e^{- 2 (- \ln (\frac{1}{5}))})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.9

Multiply $- 2 (- \ln (\frac{1}{5}))$ .

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Step 9.1.9.1

Multiply $- 1$ by $- 2$ .

$\frac{25000 (\frac{1}{5} + \frac{1}{5} - 20 e^{2 \ln (\frac{1}{5})})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.9.2

Simplify $2 \ln (\frac{1}{5})$ by moving $2$ inside the logarithm.

$\frac{25000 (\frac{1}{5} + \frac{1}{5} - 20 e^{\ln ({(\frac{1}{5})}^{2})})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.10

Exponentiation and log are inverse functions.

$\frac{25000 (\frac{1}{5} + \frac{1}{5} - 20 {(\frac{1}{5})}^{2})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.11

Apply the product rule to $\frac{1}{5}$ .

$\frac{25000 (\frac{1}{5} + \frac{1}{5} - 20 \frac{1^{2}}{5^{2}})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.12

One to any power is one.

$\frac{25000 (\frac{1}{5} + \frac{1}{5} - 20 \frac{1}{5^{2}})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.13

Raise $5$ to the power of $2$ .

$\frac{25000 (\frac{1}{5} + \frac{1}{5} - 20 (\frac{1}{25}))}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.14

Cancel the common factor of $5$ .

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Step 9.1.14.1

Factor $5$ out of $- 20$ .

$\frac{25000 (\frac{1}{5} + \frac{1}{5} + 5 (- 4) \frac{1}{25})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.14.2

Factor $5$ out of $25$ .

$\frac{25000 (\frac{1}{5} + \frac{1}{5} + 5 \cdot - 4 \frac{1}{5 \cdot 5})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.14.3

Cancel the common factor.

$\frac{25000 (\frac{1}{5} + \frac{1}{5} + 5 \cdot - 4 \frac{1}{5 \cdot 5})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.14.4

Rewrite the expression.

$\frac{25000 (\frac{1}{5} + \frac{1}{5} - 4 (\frac{1}{5}))}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.15

Combine $- 4$ and $\frac{1}{5}$ .

$\frac{25000 (\frac{1}{5} + \frac{1}{5} + \frac{- 4}{5})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.16

Move the negative in front of the fraction.

$\frac{25000 (\frac{1}{5} + \frac{1}{5} - \frac{4}{5})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.17

Combine the numerators over the common denominator.

$\frac{25000 (\frac{1 + 1}{5} - \frac{4}{5})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.18

Add $1$ and $1$ .

$\frac{25000 (\frac{2}{5} - \frac{4}{5})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.19

Combine the numerators over the common denominator.

$\frac{25000 \frac{2 - 4}{5}}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.20

Subtract $4$ from $2$ .

$\frac{25000 (\frac{- 2}{5})}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.21

Move the negative in front of the fraction.

$\frac{25000 \cdot - 1 \frac{2}{5}}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.22

Combine exponents.

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Step 9.1.22.1

Factor out negative.

$\frac{- (25000 (\frac{2}{5}))}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.22.2

Combine $25000$ and $\frac{2}{5}$ .

$\frac{- \frac{25000 \cdot 2}{5}}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.22.3

Multiply $25000$ by $2$ .

$\frac{- \frac{50000}{5}}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.1.23

Divide $50000$ by $5$ .

$\frac{- 1 \cdot 10000}{{(5 e^{- - \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.2

Simplify the denominator.

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Step 9.2.1

Multiply $- - \ln (\frac{1}{5})$ .

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Step 9.2.1.1

Multiply $- 1$ by $- 1$ .

$\frac{- 1 \cdot 10000}{{(5 e^{1 \ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.2.1.2

Multiply $\ln (\frac{1}{5})$ by $1$ .

$\frac{- 1 \cdot 10000}{{(5 e^{\ln (\frac{1}{5})} + 1)}^{4}}$

Step 9.2.2

Exponentiation and log are inverse functions.

$\frac{- 1 \cdot 10000}{{(5 (\frac{1}{5}) + 1)}^{4}}$

Step 9.2.3

Cancel the common factor of $5$ .

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Step 9.2.3.1

Cancel the common factor.

$\frac{- 1 \cdot 10000}{{(5 (\frac{1}{5}) + 1)}^{4}}$

Step 9.2.3.2

Rewrite the expression.

$\frac{- 1 \cdot 10000}{{(1 + 1)}^{4}}$

Step 9.2.4

Add $1$ and $1$ .

$\frac{- 1 \cdot 10000}{2^{4}}$

Step 9.2.5

Raise $2$ to the power of $4$ .

$\frac{- 1 \cdot 10000}{16}$

Step 9.3

Simplify the expression.

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Step 9.3.1

Multiply $- 1$ by $10000$ .

$\frac{- 10000}{16}$

Step 9.3.2

Divide $- 10000$ by $16$ .

$- 625$

Step 10

$t = - \ln (\frac{1}{5})$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$t = - \ln (\frac{1}{5})$ is a local maximum

Step 11

Find the y-value when $t = - \ln (\frac{1}{5})$ .

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Step 11.1

Replace the variable $t$ with $- \ln (\frac{1}{5})$ in the expression.

$s (- \ln (\frac{1}{5})) = \frac{25000 e^{- (- \ln (\frac{1}{5}))}}{{(1 + 5 e^{- (- \ln (\frac{1}{5}))})}^{2}}$

Step 11.2

Simplify the result.

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Step 11.2.1

Move $e^{- (- \ln (\frac{1}{5}))}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$s (- \ln (\frac{1}{5})) = \frac{25000}{{(1 + 5 e^{- (- \ln (\frac{1}{5}))})}^{2} e^{- \ln (\frac{1}{5})}}$

Step 11.2.2

Simplify the denominator.

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Step 11.2.2.1

Multiply $- - \ln (\frac{1}{5})$ .

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Step 11.2.2.1.1

Multiply $- 1$ by $- 1$ .

$s (- \ln (\frac{1}{5})) = \frac{25000}{{(1 + 5 e^{1 \ln (\frac{1}{5})})}^{2} e^{- \ln (\frac{1}{5})}}$

Step 11.2.2.1.2

Multiply $\ln (\frac{1}{5})$ by $1$ .

$s (- \ln (\frac{1}{5})) = \frac{25000}{{(1 + 5 e^{\ln (\frac{1}{5})})}^{2} e^{- \ln (\frac{1}{5})}}$

Step 11.2.2.2

Exponentiation and log are inverse functions.

$s (- \ln (\frac{1}{5})) = \frac{25000}{{(1 + 5 (\frac{1}{5}))}^{2} e^{- \ln (\frac{1}{5})}}$

Step 11.2.2.3

Cancel the common factor of $5$ .

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Step 11.2.2.3.1

Cancel the common factor.

$s (- \ln (\frac{1}{5})) = \frac{25000}{{(1 + 5 (\frac{1}{5}))}^{2} e^{- \ln (\frac{1}{5})}}$

Step 11.2.2.3.2

Rewrite the expression.

$s (- \ln (\frac{1}{5})) = \frac{25000}{{(1 + 1)}^{2} e^{- \ln (\frac{1}{5})}}$

Step 11.2.2.4

Add $1$ and $1$ .

$s (- \ln (\frac{1}{5})) = \frac{25000}{2^{2} e^{- \ln (\frac{1}{5})}}$

Step 11.2.2.5

Raise $2$ to the power of $2$ .

$s (- \ln (\frac{1}{5})) = \frac{25000}{4 e^{- \ln (\frac{1}{5})}}$

Step 11.2.2.6

Simplify $- \ln (\frac{1}{5})$ by moving $- 1$ inside the logarithm.

$s (- \ln (\frac{1}{5})) = \frac{25000}{4 e^{\ln ({(\frac{1}{5})}^{- 1})}}$

Step 11.2.2.7

Exponentiation and log are inverse functions.

$s (- \ln (\frac{1}{5})) = \frac{25000}{4 {(\frac{1}{5})}^{- 1}}$

Step 11.2.2.8

Change the sign of the exponent by rewriting the base as its reciprocal.

$s (- \ln (\frac{1}{5})) = \frac{25000}{4 \cdot 5}$

Step 11.2.3

Simplify the expression.

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Step 11.2.3.1

Multiply $4$ by $5$ .

$s (- \ln (\frac{1}{5})) = \frac{25000}{20}$

Step 11.2.3.2

Divide $25000$ by $20$ .

$s (- \ln (\frac{1}{5})) = 1250$

Step 11.2.4

The final answer is $1250$ .

$y = 1250$

Step 12

These are the local extrema for $s (t) = \frac{25000 e^{- t}}{{(1 + 5 e^{- t})}^{2}}$ .

$(- \ln (\frac{1}{5}), 1250)$ is a local maxima

Step 13