Calculus Examples

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

Step 1

Find the first derivative.

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Step 1.1

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

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

Step 1.2

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

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Step 1.2.1

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

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Step 1.2.1.1

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

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

Step 1.2.1.2

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

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

Step 1.2.1.3

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

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.2.4

Multiply $2$ by $1$ .

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

Step 1.2.5

Multiply $2$ by $- 1$ .

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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Step 1.3.2.1

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

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

Step 1.3.2.2

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

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

Step 1.3.2.3

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

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

Step 1.3.3

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

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

Step 1.3.4

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

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

Step 1.3.5

Multiply $2$ by $1$ .

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

Step 1.3.6

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

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

Step 1.3.7

Multiply $2$ by $- 1$ .

$f' (x) = - 2 \sin (2 x) - 2 \cos (2 x)$

Step 2

Find the second derivative.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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 2.2.2.1

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

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Multiply $2$ by $1$ .

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

Step 2.2.6

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

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

Step 2.2.7

Multiply $2$ by $- 2$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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 $g (x) = 2 x$ .

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Step 2.3.2.1

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

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

Step 2.3.2.2

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

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

Step 2.3.2.3

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

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

Step 2.3.3

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

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

Step 2.3.4

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

$- 4 \cos (2 x) - 2 (- \sin (2 x) (2 \cdot 1))$

Step 2.3.5

Multiply $2$ by $1$ .

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

Step 2.3.6

Multiply $2$ by $- 1$ .

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

Step 2.3.7

Multiply $- 2$ by $- 2$ .

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

Step 3

The second derivative of $f (x)$ with respect to $x$ is $- 4 \cos (2 x) + 4 \sin (2 x)$ .

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