Calculus Examples

$f (x) = \ln (\tan (5 x + 1))$

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 $g (x) = \tan (5 x + 1)$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $\tan (5 x + 1)$ .

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [\tan (5 x + 1)]$

Step 1.2

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

$\frac{1}{u_{1}} \frac{d}{d x} [\tan (5 x + 1)]$

Step 1.3

Replace all occurrences of $u_{1}$ with $\tan (5 x + 1)$ .

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

Step 2

Rewrite $\tan (5 x + 1)$ in terms of sines and cosines.

$\frac{1}{\frac{\sin (5 x + 1)}{\cos (5 x + 1)}} \frac{d}{d x} [\tan (5 x + 1)]$

Step 3

Multiply by the reciprocal of the fraction to divide by $\frac{\sin (5 x + 1)}{\cos (5 x + 1)}$ .

$\frac{\cos (5 x + 1)}{\sin (5 x + 1)} \frac{d}{d x} [\tan (5 x + 1)]$

Step 4

Convert from $\frac{\cos (5 x + 1)}{\sin (5 x + 1)}$ to $\cot (5 x + 1)$ .

$\cot (5 x + 1) \frac{d}{d x} [\tan (5 x + 1)]$

Step 5

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tan (x)$ and $g (x) = 5 x + 1$ .

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Step 5.1

To apply the Chain Rule, set $u_{2}$ as $5 x + 1$ .

$\cot (5 x + 1) (\frac{d}{d u_{2}} [\tan (u_{2})] \frac{d}{d x} [5 x + 1])$

Step 5.2

The derivative of $\tan (u_{2})$ with respect to $u_{2}$ is $\sec^{2} (u_{2})$ .

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

Step 5.3

Replace all occurrences of $u_{2}$ with $5 x + 1$ .

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

Step 6

Differentiate.

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Step 6.1

By the Sum Rule, the derivative of $5 x + 1$ with respect to $x$ is $\frac{d}{d x} [5 x] + \frac{d}{d x} [1]$ .

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

Step 6.2

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

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

Step 6.3

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

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

Step 6.4

Multiply $5$ by $1$ .

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

Step 6.5

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

$\cot (5 x + 1) \sec^{2} (5 x + 1) (5 + 0)$

Step 6.6

Simplify the expression.

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Step 6.6.1

Add $5$ and $0$ .

$\cot (5 x + 1) \sec^{2} (5 x + 1) \cdot 5$

Step 6.6.2

Move $5$ to the left of $\cot (5 x + 1) \sec^{2} (5 x + 1)$ .

$5 \cdot (\cot (5 x + 1) \sec^{2} (5 x + 1))$

Step 6.6.3

Reorder the factors of $5 \cot (5 x + 1) \sec^{2} (5 x + 1)$ .

$5 \sec^{2} (5 x + 1) \cot (5 x + 1)$