Calculus Examples

$\frac{1}{1 + \sin (x)}$

Step 1

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

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

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 $g (x) = 1 + \sin (x)$ .

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u$ with $1 + \sin (x)$ .

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

Step 3

Differentiate.

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

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

Step 4

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

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

Step 5

Simplify.

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Step 5.1

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

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

Step 5.2

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

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