Calculus Examples

$P = \frac{1}{1 + 5 e^{- \frac{1}{10} t}}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d t} (P) = \frac{d}{d t} (\frac{1}{1 + 5 e^{- \frac{1}{10} t}})$

Step 2

The derivative of $P$ with respect to $t$ is $P'$ .

$P'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Combine fractions.

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Step 3.1.1

Combine $t$ and $\frac{1}{10}$ .

$\frac{d}{d t} [\frac{1}{1 + 5 e^{- \frac{t}{10}}}]$

Step 3.1.2

Rewrite $\frac{1}{1 + 5 e^{- \frac{t}{10}}}$ as ${(1 + 5 e^{- \frac{t}{10}})}^{- 1}$ .

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

Step 3.2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{- 1}$ and $g (t) = 1 + 5 e^{- \frac{t}{10}}$ .

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Step 3.2.1

To apply the Chain Rule, set $u_{1}$ as $1 + 5 e^{- \frac{t}{10}}$ .

$\frac{d}{d u_{1}} [{u_{1}}^{- 1}] \frac{d}{d t} [1 + 5 e^{- \frac{t}{10}}]$

Step 3.2.2

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

$- {u_{1}}^{- 2} \frac{d}{d t} [1 + 5 e^{- \frac{t}{10}}]$

Step 3.2.3

Replace all occurrences of $u_{1}$ with $1 + 5 e^{- \frac{t}{10}}$ .

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

Step 3.3

Differentiate.

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Step 3.3.1

By the Sum Rule, the derivative of $1 + 5 e^{- \frac{t}{10}}$ with respect to $t$ is $\frac{d}{d t} [1] + \frac{d}{d t} [5 e^{- \frac{t}{10}}]$ .

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

Step 3.3.2

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

$- {(1 + 5 e^{- \frac{t}{10}})}^{- 2} (0 + \frac{d}{d t} [5 e^{- \frac{t}{10}}])$

Step 3.3.3

Add $0$ and $\frac{d}{d t} [5 e^{- \frac{t}{10}}]$ .

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

Step 3.3.4

Since $5$ is constant with respect to $t$ , the derivative of $5 e^{- \frac{t}{10}}$ with respect to $t$ is $5 \frac{d}{d t} [e^{- \frac{t}{10}}]$ .

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

Step 3.3.5

Multiply $5$ by $- 1$ .

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

Step 3.4

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 $g (t) = - \frac{t}{10}$ .

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Step 3.4.1

To apply the Chain Rule, set $u_{2}$ as $- \frac{t}{10}$ .

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

Step 3.4.2

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

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

Step 3.4.3

Replace all occurrences of $u_{2}$ with $- \frac{t}{10}$ .

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

Step 3.5

Differentiate.

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Step 3.5.1

Since $- \frac{1}{10}$ is constant with respect to $t$ , the derivative of $- \frac{t}{10}$ with respect to $t$ is $- \frac{1}{10} \frac{d}{d t} [t]$ .

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

Step 3.5.2

Simplify terms.

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Step 3.5.2.1

Combine $e^{- \frac{t}{10}}$ and $\frac{1}{10}$ .

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

Step 3.5.2.2

Multiply $- 1$ by $- 5$ .

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

Step 3.5.2.3

Combine $\frac{e^{- \frac{t}{10}}}{10}$ and $5$ .

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

Step 3.5.2.4

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

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

Step 3.5.2.5

Simplify the expression.

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Step 3.5.2.5.1

Move $5$ to the left of $e^{- \frac{t}{10}}$ .

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

Step 3.5.2.5.2

Move ${(1 + 5 e^{- \frac{t}{10}})}^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 3.5.2.6

Cancel the common factor of $5$ and $10$ .

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Step 3.5.2.6.1

Factor $5$ out of $5 e^{- \frac{t}{10}}$ .

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

Step 3.5.2.6.2

Cancel the common factors.

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Step 3.5.2.6.2.1

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

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

Step 3.5.2.6.2.2

Cancel the common factor.

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

Step 3.5.2.6.2.3

Rewrite the expression.

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

Step 3.5.3

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

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

Step 3.5.4

Multiply $\frac{e^{- \frac{t}{10}}}{2 {(1 + 5 e^{- \frac{t}{10}})}^{2}}$ by $1$ .

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

Step 4

Reform the equation by setting the left side equal to the right side.

$P' = \frac{e^{- \frac{t}{10}}}{2 {(1 + 5 e^{- \frac{t}{10}})}^{2}}$

Step 5

Replace $P'$ with $\frac{d P}{d t}$ .

$\frac{d P}{d t} = \frac{e^{- \frac{t}{10}}}{2 {(1 + 5 e^{- \frac{t}{10}})}^{2}}$