Calculus Examples

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

Step 1

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

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

Step 2

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

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

Multiply $- 1$ by $2$ .

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

Step 3

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

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Step 3.1

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

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Step 3.1.1

To apply the Chain Rule, set $u$ as $2 x$ .

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

Step 3.1.2

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

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

Step 3.1.3

Replace all occurrences of $u$ with $2 x$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Multiply $2$ by $1$ .

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

Step 3.5

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

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