Calculus Examples

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

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- \frac{1}{2}}$ and $g (x) = 1 - \sin (x)$ .

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Step 1.1

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

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

Step 1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = - \frac{1}{2}$ .

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

Step 1.3

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

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

Step 2

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

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

Step 3

Combine $- 1$ and $\frac{2}{2}$ .

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

Step 4

Combine the numerators over the common denominator.

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

Step 5

Simplify the numerator.

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Step 5.1

Multiply $- 1$ by $2$ .

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

Step 5.2

Subtract $2$ from $- 1$ .

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

Step 6

Differentiate.

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Step 6.1

Move the negative in front of the fraction.

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

Step 6.2

Combine fractions.

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Step 6.2.1

Combine ${(1 - \sin (x))}^{- \frac{3}{2}}$ and $\frac{1}{2}$ .

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

Step 6.2.2

Move ${(1 - \sin (x))}^{- \frac{3}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 6.3

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

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

Step 6.4

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

$- \frac{1}{2 {(1 - \sin (x))}^{\frac{3}{2}}} (0 + \frac{d}{d x} [- \sin (x)])$

Step 6.5

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

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

Step 6.6

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

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

Step 6.7

Multiply.

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Step 6.7.1

Multiply $- 1$ by $- 1$ .

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

Step 6.7.2

Multiply $\frac{1}{2 {(1 - \sin (x))}^{\frac{3}{2}}}$ by $1$ .

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

Step 7

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

$\frac{1}{2 {(1 - \sin (x))}^{\frac{3}{2}}} \cos (x)$

Step 8

Combine $\frac{1}{2 {(1 - \sin (x))}^{\frac{3}{2}}}$ and $\cos (x)$ .

$\frac{\cos (x)}{2 {(1 - \sin (x))}^{\frac{3}{2}}}$