Calculus Examples

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

Step 1

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

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

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

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u_{1}$ with $\sin (x + 1)$ .

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

Step 3

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

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

Step 4

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

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

Step 5

Combine the numerators over the common denominator.

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

Step 6

Simplify the numerator.

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Step 6.1

Multiply $- 1$ by $2$ .

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

Step 6.2

Subtract $2$ from $1$ .

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

Step 7

Combine fractions.

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Step 7.1

Move the negative in front of the fraction.

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

Step 7.2

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

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

Step 7.3

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

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

Step 8

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

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Step 8.1

To apply the Chain Rule, set $u_{2}$ as $x + 1$ .

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

Step 8.2

The derivative of $\sin (u_{2})$ with respect to $u_{2}$ is $\cos (u_{2})$ .

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

Step 8.3

Replace all occurrences of $u_{2}$ with $x + 1$ .

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

Step 9

Differentiate.

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Step 9.1

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

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

Step 9.2

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

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

Step 9.3

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

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

Step 9.4

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

$\frac{\cos (x + 1)}{2 {\sin (x + 1)}^{\frac{1}{2}}} (1 + 0)$

Step 9.5

Simplify the expression.

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Step 9.5.1

Add $1$ and $0$ .

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

Step 9.5.2

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

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

Step 10

Simplify.

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Step 10.1

Separate fractions.

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

Step 10.2

Convert from $\frac{1}{{\sin (x + 1)}^{\frac{1}{2}}}$ to ${\csc (x + 1)}^{\frac{1}{2}}$ .

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

Step 10.3

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

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

Step 10.4

Reorder factors in $\frac{\cos (x + 1) {\csc (x + 1)}^{\frac{1}{2}}}{2}$ .

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