Calculus Examples

$S (t) = 68 - 20 \ln (t + 1)$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of $68 - 20 \ln (t + 1)$ with respect to $t$ is $\frac{d}{d t} [68] + \frac{d}{d t} [- 20 \ln (t + 1)]$ .

$\frac{d}{d t} [68] + \frac{d}{d t} [- 20 \ln (t + 1)]$

Step 1.1.2

Since $68$ is constant with respect to $t$ , the derivative of $68$ with respect to $t$ is $0$ .

$0 + \frac{d}{d t} [- 20 \ln (t + 1)]$

Step 1.2

Evaluate $\frac{d}{d t} [- 20 \ln (t + 1)]$ .

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Step 1.2.1

Since $- 20$ is constant with respect to $t$ , the derivative of $- 20 \ln (t + 1)$ with respect to $t$ is $- 20 \frac{d}{d t} [\ln (t + 1)]$ .

$0 - 20 \frac{d}{d t} [\ln (t + 1)]$

Step 1.2.2

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

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Step 1.2.2.1

To apply the Chain Rule, set $u$ as $t + 1$ .

$0 - 20 (\frac{d}{d u} [\ln (u)] \frac{d}{d t} [t + 1])$

Step 1.2.2.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$0 - 20 (\frac{1}{u} \frac{d}{d t} [t + 1])$

Step 1.2.2.3

Replace all occurrences of $u$ with $t + 1$ .

$0 - 20 (\frac{1}{t + 1} \frac{d}{d t} [t + 1])$

Step 1.2.3

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

$0 - 20 (\frac{1}{t + 1} (\frac{d}{d t} [t] + \frac{d}{d t} [1]))$

Step 1.2.4

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

$0 - 20 (\frac{1}{t + 1} (1 + \frac{d}{d t} [1]))$

Step 1.2.5

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

$0 - 20 (\frac{1}{t + 1} (1 + 0))$

Step 1.2.6

Add $1$ and $0$ .

$0 - 20 (\frac{1}{t + 1} \cdot 1)$

Step 1.2.7

Multiply $\frac{1}{t + 1}$ by $1$ .

$0 - 20 \frac{1}{t + 1}$

Step 1.2.8

Combine $- 20$ and $\frac{1}{t + 1}$ .

$0 + \frac{- 20}{t + 1}$

Step 1.2.9

Move the negative in front of the fraction.

$0 - \frac{20}{t + 1}$

Step 1.3

Subtract $\frac{20}{t + 1}$ from $0$ .

$- \frac{20}{t + 1}$

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate using the Constant Multiple Rule.

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Step 2.1.1

Since $- 20$ is constant with respect to $t$ , the derivative of $- \frac{20}{t + 1}$ with respect to $t$ is $- 20 \frac{d}{d t} [\frac{1}{t + 1}]$ .

$S'' (t) = - 20 \frac{d}{d t} \frac{1}{t + 1}$

Step 2.1.2

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

$S'' (t) = - 20 \frac{d}{d t} {(t + 1)}^{- 1}$

Step 2.2

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 $g (t) = t + 1$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $t + 1$ .

$S'' (t) = - 20 (\frac{d}{d u} (u^{- 1}) \frac{d}{d t} (t + 1))$

Step 2.2.2

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

$S'' (t) = - 20 (- u^{- 2} \frac{d}{d t} (t + 1))$

Step 2.2.3

Replace all occurrences of $u$ with $t + 1$ .

$S'' (t) = - 20 (- {(t + 1)}^{- 2} \frac{d}{d t} (t + 1))$

Step 2.3

Differentiate.

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Step 2.3.1

Multiply $- 1$ by $- 20$ .

$S'' (t) = 20 ({(t + 1)}^{- 2} \frac{d}{d t} (t + 1))$

Step 2.3.2

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

$S'' (t) = 20 {(t + 1)}^{- 2} (\frac{d}{d t} (t) + \frac{d}{d t} (1))$

Step 2.3.3

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

$S'' (t) = 20 {(t + 1)}^{- 2} (1 + \frac{d}{d t} (1))$

Step 2.3.4

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

$S'' (t) = 20 {(t + 1)}^{- 2} (1 + 0)$

Step 2.3.5

Simplify the expression.

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Step 2.3.5.1

Add $1$ and $0$ .

$S'' (t) = 20 {(t + 1)}^{- 2} \cdot 1$

Step 2.3.5.2

Multiply $20$ by $1$ .

$S'' (t) = 20 {(t + 1)}^{- 2}$

Step 2.4

Simplify.

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Step 2.4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$S'' (t) = 20 (\frac{1}{{(t + 1)}^{2}})$

Step 2.4.2

Combine $20$ and $\frac{1}{{(t + 1)}^{2}}$ .

$S'' (t) = \frac{20}{{(t + 1)}^{2}}$

Step 3

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

$- \frac{20}{t + 1} = 0$

Step 4

Since there is no value of $x$ that makes the first derivative equal to $0$ , there are no local extrema.

No Local Extrema

Step 5

No Local Extrema