Calculus Examples

$P (t) = (t^{2} + 100) \ln (t + 2)$

Step 1

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 $f (t) = t^{2} + 100$ and $g (t) = \ln (t + 2)$ .

$(t^{2} + 100) \frac{d}{d t} [\ln (t + 2)] + \ln (t + 2) \frac{d}{d t} [t^{2} + 100]$

Step 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 $g (t) = t + 2$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $t + 2$ .

$(t^{2} + 100) (\frac{d}{d u} [\ln (u)] \frac{d}{d t} [t + 2]) + \ln (t + 2) \frac{d}{d t} [t^{2} + 100]$

Step 2.2

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

$(t^{2} + 100) (\frac{1}{u} \frac{d}{d t} [t + 2]) + \ln (t + 2) \frac{d}{d t} [t^{2} + 100]$

Step 2.3

Replace all occurrences of $u$ with $t + 2$ .

$(t^{2} + 100) (\frac{1}{t + 2} \frac{d}{d t} [t + 2]) + \ln (t + 2) \frac{d}{d t} [t^{2} + 100]$

Step 3

Differentiate.

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Step 3.1

By the Sum Rule, the derivative of $t + 2$ with respect to $t$ is $\frac{d}{d t} [t] + \frac{d}{d t} [2]$ .

$(t^{2} + 100) \frac{1}{t + 2} (\frac{d}{d t} [t] + \frac{d}{d t} [2]) + \ln (t + 2) \frac{d}{d t} [t^{2} + 100]$

Step 3.2

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

$(t^{2} + 100) \frac{1}{t + 2} (1 + \frac{d}{d t} [2]) + \ln (t + 2) \frac{d}{d t} [t^{2} + 100]$

Step 3.3

Since $2$ is constant with respect to $t$ , the derivative of $2$ with respect to $t$ is $0$ .

$(t^{2} + 100) \frac{1}{t + 2} (1 + 0) + \ln (t + 2) \frac{d}{d t} [t^{2} + 100]$

Step 3.4

Simplify the expression.

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Step 3.4.1

Add $1$ and $0$ .

$(t^{2} + 100) \frac{1}{t + 2} \cdot 1 + \ln (t + 2) \frac{d}{d t} [t^{2} + 100]$

Step 3.4.2

Multiply $t^{2} + 100$ by $1$ .

$(t^{2} + 100) \frac{1}{t + 2} + \ln (t + 2) \frac{d}{d t} [t^{2} + 100]$

Step 3.5

By the Sum Rule, the derivative of $t^{2} + 100$ with respect to $t$ is $\frac{d}{d t} [t^{2}] + \frac{d}{d t} [100]$ .

$(t^{2} + 100) \frac{1}{t + 2} + \ln (t + 2) (\frac{d}{d t} [t^{2}] + \frac{d}{d t} [100])$

Step 3.6

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

$(t^{2} + 100) \frac{1}{t + 2} + \ln (t + 2) (2 t + \frac{d}{d t} [100])$

Step 3.7

Since $100$ is constant with respect to $t$ , the derivative of $100$ with respect to $t$ is $0$ .

$(t^{2} + 100) \frac{1}{t + 2} + \ln (t + 2) (2 t + 0)$

Step 3.8

Simplify the expression.

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Step 3.8.1

Add $2 t$ and $0$ .

$(t^{2} + 100) \frac{1}{t + 2} + \ln (t + 2) (2 t)$

Step 3.8.2

Reorder terms.

$(t^{2} + 100) \frac{1}{t + 2} + 2 t \ln (t + 2)$