Find the Derivative - d/d@VAR f(x)=sin(x^2-12)

$f (x) = \sin (x^{2} - 12)$

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 $g (x) = x^{2} - 12$ .

To apply the Chain Rule, set $u$ as $x^{2} - 12$ .

$\frac{d}{d u} [\sin (u)] \frac{d}{d x} [x^{2} - 12]$

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

$\cos (u) \frac{d}{d x} [x^{2} - 12]$

Replace all occurrences of $u$ with $x^{2} - 12$ .

$\cos (x^{2} - 12) \frac{d}{d x} [x^{2} - 12]$

Differentiate.

By the Sum Rule, the derivative of $x^{2} - 12$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 12]$ .

$\cos (x^{2} - 12) (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 12])$

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

$\cos (x^{2} - 12) (2 x + \frac{d}{d x} [- 12])$

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12$ with respect to $x$ is $0$ .

$\cos (x^{2} - 12) (2 x + 0)$

Simplify the expression.

Add $2 x$ and $0$ .

$\cos (x^{2} - 12) (2 x)$

Reorder the factors of $\cos (x^{2} - 12) (2 x)$ .

$2 x \cos (x^{2} - 12)$

Find the Derivative - d/d@VAR f(x)=sin(x^2-12)

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