Calculus Examples

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

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{1 + \sin (x)}$ as ${(1 + \sin (x))}^{\frac{1}{2}}$ .

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

Step 2

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

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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 3

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 4

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 5

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 6

Simplify the numerator.

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Step 6.1

Multiply $- 1$ by $2$ .

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

Step 6.2

Subtract $2$ from $1$ .

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

Step 7

Differentiate.

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Step 7.1

Move the negative in front of the fraction.

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

Step 7.2

Combine fractions.

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Step 7.2.1

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

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

Step 7.2.2

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

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

Step 7.3

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

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

Step 7.4

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

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

Step 7.5

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

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

Step 8

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

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

Step 9

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

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