Calculus Examples

$y = \sqrt{2 + \cos (2 x)}$

Step 1

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

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

Step 2

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} ({(2 + \cos (2 x))}^{\frac{1}{2}})$

Step 3

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 4

Differentiate the right side of the equation.

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Step 4.1

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 $g (x) = 2 + \cos (2 x)$ .

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Step 4.1.1

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

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

Step 4.1.2

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

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

Step 4.1.3

Replace all occurrences of $u_{1}$ with $2 + \cos (2 x)$ .

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

Step 4.2

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

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

Step 4.3

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

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

Step 4.4

Combine the numerators over the common denominator.

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

Step 4.5

Simplify the numerator.

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Step 4.5.1

Multiply $- 1$ by $2$ .

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

Step 4.5.2

Subtract $2$ from $1$ .

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

Step 4.6

Differentiate.

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Step 4.6.1

Move the negative in front of the fraction.

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

Step 4.6.2

Combine fractions.

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Step 4.6.2.1

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

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

Step 4.6.2.2

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

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

Step 4.6.3

By the Sum Rule, the derivative of $2 + \cos (2 x)$ with respect to $x$ is $\frac{d}{d x} [2] + \frac{d}{d x} [\cos (2 x)]$ .

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

Step 4.6.4

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

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

Step 4.6.5

Add $0$ and $\frac{d}{d x} [\cos (2 x)]$ .

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

Step 4.7

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

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Step 4.7.1

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

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

Step 4.7.2

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

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

Step 4.7.3

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

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

Step 4.8

Differentiate using the Constant Multiple Rule.

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Step 4.8.1

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

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

Step 4.8.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 4.8.3

Simplify terms.

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Step 4.8.3.1

Multiply $2$ by $- 1$ .

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

Step 4.8.3.2

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

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

Step 4.8.3.3

Factor $2$ out of $- 2 \sin (2 x)$ .

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

Step 4.9

Cancel the common factors.

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Step 4.9.1

Factor $2$ out of $2 {(2 + \cos (2 x))}^{\frac{1}{2}}$ .

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

Step 4.9.2

Cancel the common factor.

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

Step 4.9.3

Rewrite the expression.

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

Step 4.10

Move the negative in front of the fraction.

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

Step 4.11

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

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

Step 4.12

Multiply $- 1$ by $1$ .

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

Step 5

Reform the equation by setting the left side equal to the right side.

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

Step 6

Replace $y'$ with $\frac{d y}{d x}$ .

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