Calculus Examples

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

Step 1

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

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

Step 2

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

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Step 2.1

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

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

Step 2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = \cos (x)$ and $g (x) = \sin (2 x)$ .

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

Step 2.3

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 2.3.1

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

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

Step 2.3.2

The derivative of $\sin (u_{1})$ with respect to $u_{1}$ is $\cos (u_{1})$ .

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

Step 2.3.3

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

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

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

Step 2.7

Multiply $2$ by $1$ .

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

Step 2.8

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

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

Step 2.9

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

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

Step 3

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

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Step 3.1

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

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

Step 3.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = \sin (x)$ and $g (x) = \cos (2 x)$ .

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

Step 3.3

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 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.4

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

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

Step 3.5

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

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

Step 3.6

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

$2 (2 \cos (x) \cos (2 x) + \sin (2 x) (- \sin (x))) - (\sin (x) (- \sin (2 x) (2 \cdot 1)) + \cos (2 x) \cos (x))$

Step 3.7

Multiply $2$ by $1$ .

$2 (2 \cos (x) \cos (2 x) + \sin (2 x) (- \sin (x))) - (\sin (x) (- \sin (2 x) \cdot 2) + \cos (2 x) \cos (x))$

Step 3.8

Multiply $2$ by $- 1$ .

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

Step 4

Simplify.

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Step 4.1

Apply the distributive property.

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

Step 4.2

Apply the distributive property.

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

Step 4.3

Combine terms.

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Step 4.3.1

Multiply $2$ by $2$ .

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

Step 4.3.2

Multiply $- 1$ by $2$ .

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

Step 4.3.3

Multiply $- 2$ by $- 1$ .

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

Step 4.3.4

Reorder the factors of $- 2 \sin (2 x) \sin (x)$ .

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

Step 4.3.5

Add $- 2 \sin (x) \sin (2 x)$ and $2 \sin (x) \sin (2 x)$ .

$4 \cos (x) \cos (2 x) + 0 - \cos (2 x) \cos (x)$

Step 4.3.6

Add $4 \cos (x) \cos (2 x)$ and $0$ .

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

Step 4.3.7

Reorder the factors of $- \cos (2 x) \cos (x)$ .

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

Step 4.3.8

Subtract $\cos (x) \cos (2 x)$ from $4 \cos (x) \cos (2 x)$ .

$3 \cos (x) \cos (2 x)$