Calculus Examples

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

Step 1

Differentiate.

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Step 1.1

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

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

Step 1.2

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

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

Step 2

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

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Multiply $2$ by $1$ .

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

Step 2.6

Multiply $2$ by $- 1$ .

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

Step 2.7

Multiply $- 2$ by $- 1$ .

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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

$0 + 2 \sin (2 x) + 2 (2 u_{2} \frac{d}{d x} [\cos (x)])$

Step 3.2.3

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

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

Step 3.3

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

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

Step 3.4

Multiply $- 1$ by $2$ .

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

Step 3.5

Multiply $- 2$ by $2$ .

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

Step 4

Simplify.

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Step 4.1

Add $0$ and $2 \sin (2 x)$ .

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

Step 4.2

Reorder terms.

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