Calculus Examples

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

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

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Step 1.1

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

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

Step 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 $n = \frac{1}{2}$ .

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

Step 1.3

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

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

Step 2

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

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

Step 3

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

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

Step 4

Combine the numerators over the common denominator.

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

Step 5

Simplify the numerator.

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Step 5.1

Multiply $- 1$ by $2$ .

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

Step 5.2

Subtract $2$ from $1$ .

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

Step 6

Combine fractions.

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Step 6.1

Move the negative in front of the fraction.

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

Step 6.2

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

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

Step 6.3

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

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

Step 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 7.1

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

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

Step 7.2

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

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

Step 7.3

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

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

Step 8

Differentiate using the Constant Multiple Rule.

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Step 8.1

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

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

Step 8.2

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

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

Step 8.3

Simplify terms.

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Step 8.3.1

Multiply $2$ by $- 1$ .

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

Step 8.3.2

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

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

Step 8.3.3

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

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

Step 9

Cancel the common factors.

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Step 9.1

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

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

Step 9.2

Cancel the common factor.

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

Step 9.3

Rewrite the expression.

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

Step 10

Move the negative in front of the fraction.

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

Step 11

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

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

Step 12

Multiply $- 1$ by $1$ .

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

Step 13

Simplify.

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Step 13.1

Separate fractions.

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

Step 13.2

Convert from $\frac{1}{{\cos (2 x)}^{\frac{1}{2}}}$ to ${\sec (2 x)}^{\frac{1}{2}}$ .

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

Step 13.3

Divide $\sin (2 x)$ by $1$ .

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

Step 13.4

Reorder factors in $- \sin (2 x) {\sec (2 x)}^{\frac{1}{2}}$ .

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