Calculus Examples

$f (x) = \ln (e^{x^{3}}) + 6 \ln (x - 1)$

Step 1

By the Sum Rule, the derivative of $\ln (e^{x^{3}}) + 6 \ln (x - 1)$ with respect to $x$ is $\frac{d}{d x} [\ln (e^{x^{3}})] + \frac{d}{d x} [6 \ln (x - 1)]$ .

$\frac{d}{d x} [\ln (e^{x^{3}})] + \frac{d}{d x} [6 \ln (x - 1)]$

Step 2

Evaluate $\frac{d}{d x} [\ln (e^{x^{3}})]$ .

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Step 2.1

Use logarithm rules to move $x^{3}$ out of the exponent.

$\frac{d}{d x} [x^{3} \ln (e)] + \frac{d}{d x} [6 \ln (x - 1)]$

Step 2.2

The natural logarithm of $e$ is $1$ .

$\frac{d}{d x} [x^{3} \cdot 1] + \frac{d}{d x} [6 \ln (x - 1)]$

Step 2.3

Multiply $x^{3}$ by $1$ .

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [6 \ln (x - 1)]$

Step 2.4

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

$3 x^{2} + \frac{d}{d x} [6 \ln (x - 1)]$

Step 3

Evaluate $\frac{d}{d x} [6 \ln (x - 1)]$ .

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Step 3.1

Since $6$ is constant with respect to $x$ , the derivative of $6 \ln (x - 1)$ with respect to $x$ is $6 \frac{d}{d x} [\ln (x - 1)]$ .

$3 x^{2} + 6 \frac{d}{d x} [\ln (x - 1)]$

Step 3.2

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 $g (x) = x - 1$ .

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Step 3.2.1

To apply the Chain Rule, set $u$ as $x - 1$ .

$3 x^{2} + 6 (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x - 1])$

Step 3.2.2

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

$3 x^{2} + 6 (\frac{1}{u} \frac{d}{d x} [x - 1])$

Step 3.2.3

Replace all occurrences of $u$ with $x - 1$ .

$3 x^{2} + 6 (\frac{1}{x - 1} \frac{d}{d x} [x - 1])$

Step 3.3

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

$3 x^{2} + 6 (\frac{1}{x - 1} (\frac{d}{d x} [x] + \frac{d}{d x} [- 1]))$

Step 3.4

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

$3 x^{2} + 6 (\frac{1}{x - 1} (1 + \frac{d}{d x} [- 1]))$

Step 3.5

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

$3 x^{2} + 6 (\frac{1}{x - 1} (1 + 0))$

Step 3.6

Add $1$ and $0$ .

$3 x^{2} + 6 (\frac{1}{x - 1} \cdot 1)$

Step 3.7

Multiply $\frac{1}{x - 1}$ by $1$ .

$3 x^{2} + 6 \frac{1}{x - 1}$

Step 3.8

Combine $6$ and $\frac{1}{x - 1}$ .

$3 x^{2} + \frac{6}{x - 1}$

Step 4

Combine terms.

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Step 4.1

To write $3 x^{2}$ as a fraction with a common denominator, multiply by $\frac{x - 1}{x - 1}$ .

$\frac{3 x^{2} (x - 1)}{x - 1} + \frac{6}{x - 1}$

Step 4.2

Combine the numerators over the common denominator.

$\frac{3 x^{2} (x - 1) + 6}{x - 1}$