Calculus Examples

$f (x) = \frac{\tan (x - 5)}{x - 1}$

Step 1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = \tan (x - 5)$ and $g (x) = x - 1$ .

$\frac{(x - 1) \frac{d}{d x} [\tan (x - 5)] - \tan (x - 5) \frac{d}{d x} [x - 1]}{{(x - 1)}^{2}}$

Step 2

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

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Step 2.1

To apply the Chain Rule, set $u$ as $x - 5$ .

$\frac{(x - 1) (\frac{d}{d u} [\tan (u)] \frac{d}{d x} [x - 5]) - \tan (x - 5) \frac{d}{d x} [x - 1]}{{(x - 1)}^{2}}$

Step 2.2

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

$\frac{(x - 1) (\sec^{2} (u) \frac{d}{d x} [x - 5]) - \tan (x - 5) \frac{d}{d x} [x - 1]}{{(x - 1)}^{2}}$

Step 2.3

Replace all occurrences of $u$ with $x - 5$ .

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

Step 3

Differentiate.

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

$\frac{(x - 1) \sec^{2} (x - 5) (1 + 0) - \tan (x - 5) \frac{d}{d x} [x - 1]}{{(x - 1)}^{2}}$

Step 3.4

Simplify the expression.

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Step 3.4.1

Add $1$ and $0$ .

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

Step 3.4.2

Multiply $x - 1$ by $1$ .

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

Step 3.5

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

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

Step 3.6

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

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

Step 3.7

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

$\frac{(x - 1) \sec^{2} (x - 5) - \tan (x - 5) (1 + 0)}{{(x - 1)}^{2}}$

Step 3.8

Simplify terms.

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Step 3.8.1

Add $1$ and $0$ .

$\frac{(x - 1) \sec^{2} (x - 5) - \tan (x - 5) \cdot 1}{{(x - 1)}^{2}}$

Step 3.8.2

Multiply $- 1$ by $1$ .

$\frac{(x - 1) \sec^{2} (x - 5) - \tan (x - 5)}{{(x - 1)}^{2}}$

Step 3.8.3

Simplify each term.

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Step 3.8.3.1

Apply the distributive property.

$\frac{x \sec^{2} (x - 5) - 1 \sec^{2} (x - 5) - \tan (x - 5)}{{(x - 1)}^{2}}$

Step 3.8.3.2

Rewrite $- 1 \sec^{2} (x - 5)$ as $- \sec^{2} (x - 5)$ .

$\frac{x \sec^{2} (x - 5) - \sec^{2} (x - 5) - \tan (x - 5)}{{(x - 1)}^{2}}$