Find the Derivative - d/dx 1+cos(18x)

$1 + \cos (18 x)$

Step 1

Differentiate.

By the Sum Rule, the derivative of $1 + \cos (18 x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\cos (18 x)]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [\cos (18 x)]$

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

$0 + \frac{d}{d x} [\cos (18 x)]$

Step 2

Evaluate $\frac{d}{d x} [\cos (18 x)]$ .

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 $g (x) = 18 x$ .

To apply the Chain Rule, set $u$ as $18 x$ .

$0 + \frac{d}{d u} [\cos (u)] \frac{d}{d x} [18 x]$

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

$0 - \sin (u) \frac{d}{d x} [18 x]$

Replace all occurrences of $u$ with $18 x$ .

$0 - \sin (18 x) \frac{d}{d x} [18 x]$

Since $18$ is constant with respect to $x$ , the derivative of $18 x$ with respect to $x$ is $18 \frac{d}{d x} [x]$ .

$0 - \sin (18 x) (18 \frac{d}{d x} [x])$

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

$0 - \sin (18 x) (18 \cdot 1)$

Multiply $18$ by $1$ .

$0 - \sin (18 x) \cdot 18$

Multiply $18$ by $- 1$ .

$0 - 18 \sin (18 x)$

Step 3

Subtract $18 \sin (18 x)$ from $0$ .

$- 18 \sin (18 x)$

Find the Derivative - d/dx 1+cos(18x)

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