Calculus Examples

$\frac{1}{1 - \cos (x)}$

Step 1

Rewrite $\frac{1}{1 - \cos (x)}$ as ${(1 - \cos (x))}^{- 1}$ .

$\frac{d}{d x} [{(1 - \cos (x))}^{- 1}]$

Step 2

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

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Step 2.1

To apply the Chain Rule, set $u$ as $1 - \cos (x)$ .

$\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [1 - \cos (x)]$

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u$ with $1 - \cos (x)$ .

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

Step 3

Differentiate.

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Add $0$ and $\frac{d}{d x} [- \cos (x)]$ .

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

Step 3.4

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

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

Step 3.5

Multiply.

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Step 3.5.1

Multiply $- 1$ by $- 1$ .

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

Step 3.5.2

Multiply ${(1 - \cos (x))}^{- 2}$ by $1$ .

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

Step 4

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

${(1 - \cos (x))}^{- 2} (- \sin (x))$

Step 5

Simplify.

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Step 5.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{{(1 - \cos (x))}^{2}} (- \sin (x))$

Step 5.2

Combine $\frac{1}{{(1 - \cos (x))}^{2}}$ and $\sin (x)$ .

$- \frac{\sin (x)}{{(1 - \cos (x))}^{2}}$