Find the Derivative minus d First, divide dx sin(three x)

$\sin (3 x)$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = 3 x$ .

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To apply the Chain Rule, set $u$ as $3 x$ .

$\frac{d}{d u} [\sin (u)] \frac{d}{d x} [3 x]$

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

$\cos (u) \frac{d}{d x} [3 x]$

Replace all occurrences of $u$ with $3 x$ .

$\cos (3 x) \frac{d}{d x} [3 x]$

Differentiate.

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Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$\cos (3 x) (3 \frac{d}{d x} [x])$

Put $3$ to the left of $\cos (3 x)$ .

$3 \cdot \cos (3 x) \frac{d}{d x} [x]$

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

$3 \cos (3 x) \cdot 1$

First, multiply $3$ by $1$ .

$3 \cos (3 x)$

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