Calculus Examples

$f (x) = \ln (x^{3} + 5)$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = x^{3} + 5$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $x^{3} + 5$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x^{3} + 5]$

Step 1.2

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

$\frac{1}{u} \frac{d}{d x} [x^{3} + 5]$

Step 1.3

Replace all occurrences of $u$ with $x^{3} + 5$ .

$\frac{1}{x^{3} + 5} \frac{d}{d x} [x^{3} + 5]$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $x^{3} + 5$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [5]$ .

$\frac{1}{x^{3} + 5} (\frac{d}{d x} [x^{3}] + \frac{d}{d x} [5])$

Step 2.2

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

$\frac{1}{x^{3} + 5} (3 x^{2} + \frac{d}{d x} [5])$

Step 2.3

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$\frac{1}{x^{3} + 5} (3 x^{2} + 0)$

Step 2.4

Combine fractions.

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Step 2.4.1

Add $3 x^{2}$ and $0$ .

$\frac{1}{x^{3} + 5} (3 x^{2})$

Step 2.4.2

Combine $3$ and $\frac{1}{x^{3} + 5}$ .

$\frac{3}{x^{3} + 5} x^{2}$

Step 2.4.3

Combine $\frac{3}{x^{3} + 5}$ and $x^{2}$ .

$\frac{3 x^{2}}{x^{3} + 5}$