Precalculus Examples

$y = \sin^{2} (2 x) + \frac{1}{2} \cos (4 x)$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (\sin^{2} (2 x) + \frac{1}{2} \cdot \cos (4 x))$

Step 2

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

$y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $\sin^{2} (2 x) + \frac{1}{2} \cdot \cos (4 x)$ with respect to $x$ is $\frac{d}{d x} [\sin^{2} (2 x)] + \frac{d}{d x} [\frac{1}{2} \cdot \cos (4 x)]$ .

$\frac{d}{d x} [\sin^{2} (2 x)] + \frac{d}{d x} [\frac{1}{2} \cdot \cos (4 x)]$

Step 3.2

Evaluate $\frac{d}{d x} [\sin^{2} (2 x)]$ .

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Step 3.2.1

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

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Step 3.2.1.1

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

$\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [\sin (2 x)] + \frac{d}{d x} [\frac{1}{2} \cdot \cos (4 x)]$

Step 3.2.1.2

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

$2 u_{1} \frac{d}{d x} [\sin (2 x)] + \frac{d}{d x} [\frac{1}{2} \cdot \cos (4 x)]$

Step 3.2.1.3

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

$2 \sin (2 x) \frac{d}{d x} [\sin (2 x)] + \frac{d}{d x} [\frac{1}{2} \cdot \cos (4 x)]$

Step 3.2.2

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

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Step 3.2.2.1

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

$2 \sin (2 x) (\frac{d}{d u_{2}} [\sin (u_{2})] \frac{d}{d x} [2 x]) + \frac{d}{d x} [\frac{1}{2} \cdot \cos (4 x)]$

Step 3.2.2.2

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

$2 \sin (2 x) (\cos (u_{2}) \frac{d}{d x} [2 x]) + \frac{d}{d x} [\frac{1}{2} \cdot \cos (4 x)]$

Step 3.2.2.3

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

$2 \sin (2 x) (\cos (2 x) \frac{d}{d x} [2 x]) + \frac{d}{d x} [\frac{1}{2} \cdot \cos (4 x)]$

Step 3.2.3

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

$2 \sin (2 x) (\cos (2 x) (2 \frac{d}{d x} [x])) + \frac{d}{d x} [\frac{1}{2} \cdot \cos (4 x)]$

Step 3.2.4

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

$2 \sin (2 x) (\cos (2 x) (2 \cdot 1)) + \frac{d}{d x} [\frac{1}{2} \cdot \cos (4 x)]$

Step 3.2.5

Multiply $2$ by $1$ .

$2 \sin (2 x) (\cos (2 x) \cdot 2) + \frac{d}{d x} [\frac{1}{2} \cdot \cos (4 x)]$

Step 3.2.6

Move $2$ to the left of $\cos (2 x)$ .

$2 \sin (2 x) (2 \cdot \cos (2 x)) + \frac{d}{d x} [\frac{1}{2} \cdot \cos (4 x)]$

Step 3.2.7

Multiply $2$ by $2$ .

$4 \sin (2 x) \cos (2 x) + \frac{d}{d x} [\frac{1}{2} \cdot \cos (4 x)]$

Step 3.3

Evaluate $\frac{d}{d x} [\frac{1}{2} \cdot \cos (4 x)]$ .

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Step 3.3.1

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{1}{2} \cdot \cos (4 x)$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [\cos (4 x)]$ .

$4 \sin (2 x) \cos (2 x) + \frac{1}{2} \frac{d}{d x} [\cos (4 x)]$

Step 3.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = 4 x$ .

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Step 3.3.2.1

To apply the Chain Rule, set $u_{3}$ as $4 x$ .

$4 \sin (2 x) \cos (2 x) + \frac{1}{2} (\frac{d}{d u_{3}} [\cos (u_{3})] \frac{d}{d x} [4 x])$

Step 3.3.2.2

The derivative of $\cos (u_{3})$ with respect to $u_{3}$ is $- \sin (u_{3})$ .

$4 \sin (2 x) \cos (2 x) + \frac{1}{2} (- \sin (u_{3}) \frac{d}{d x} [4 x])$

Step 3.3.2.3

Replace all occurrences of $u_{3}$ with $4 x$ .

$4 \sin (2 x) \cos (2 x) + \frac{1}{2} (- \sin (4 x) \frac{d}{d x} [4 x])$

Step 3.3.3

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$4 \sin (2 x) \cos (2 x) + \frac{1}{2} (- \sin (4 x) (4 \frac{d}{d x} [x]))$

Step 3.3.4

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

$4 \sin (2 x) \cos (2 x) + \frac{1}{2} (- \sin (4 x) (4 \cdot 1))$

Step 3.3.5

Multiply $4$ by $1$ .

$4 \sin (2 x) \cos (2 x) + \frac{1}{2} (- \sin (4 x) \cdot 4)$

Step 3.3.6

Multiply $4$ by $- 1$ .

$4 \sin (2 x) \cos (2 x) + \frac{1}{2} (- 4 \sin (4 x))$

Step 3.3.7

Combine $- 4$ and $\frac{1}{2}$ .

$4 \sin (2 x) \cos (2 x) + \frac{- 4}{2} \sin (4 x)$

Step 3.3.8

Combine $\frac{- 4}{2}$ and $\sin (4 x)$ .

$4 \sin (2 x) \cos (2 x) + \frac{- 4 \sin (4 x)}{2}$

Step 3.3.9

Cancel the common factor of $- 4$ and $2$ .

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Step 3.3.9.1

Factor $2$ out of $- 4 \sin (4 x)$ .

$4 \sin (2 x) \cos (2 x) + \frac{2 (- 2 \sin (4 x))}{2}$

Step 3.3.9.2

Cancel the common factors.

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Step 3.3.9.2.1

Factor $2$ out of $2$ .

$4 \sin (2 x) \cos (2 x) + \frac{2 (- 2 \sin (4 x))}{2 (1)}$

Step 3.3.9.2.2

Cancel the common factor.

$4 \sin (2 x) \cos (2 x) + \frac{2 (- 2 \sin (4 x))}{2 \cdot 1}$

Step 3.3.9.2.3

Rewrite the expression.

$4 \sin (2 x) \cos (2 x) + \frac{- 2 \sin (4 x)}{1}$

Step 3.3.9.2.4

Divide $- 2 \sin (4 x)$ by $1$ .

$4 \sin (2 x) \cos (2 x) - 2 \sin (4 x)$

Step 3.4

Reorder terms.

$4 \cos (2 x) \sin (2 x) - 2 \sin (4 x)$

Step 4

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

$y' = 4 \cos (2 x) \sin (2 x) - 2 \sin (4 x)$

Step 5

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

$\frac{d y}{d x} = 4 \cos (2 x) \sin (2 x) - 2 \sin (4 x)$