Calculus Examples

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

Step 1

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 $g (t) = t^{2} + 1$ .

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To apply the Chain Rule, set $u$ as $t^{2} + 1$ .

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

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

$\frac{1}{u} \frac{d}{d t} [t^{2} + 1]$

Replace all occurrences of $u$ with $t^{2} + 1$ .

$\frac{1}{t^{2} + 1} \frac{d}{d t} [t^{2} + 1]$

Step 2

Differentiate.

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By the Sum Rule, the derivative of $t^{2} + 1$ with respect to $t$ is $\frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$ .

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

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

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

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

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

Combine fractions.

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Add $2 t$ and $0$ .

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

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

$\frac{2}{t^{2} + 1} t$

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

$\frac{2 t}{t^{2} + 1}$