Find the Derivative – d/dx cos(x^2-3x)

$\cos (x^{2} - 3 x)$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = x^{2} - 3 x$ .

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To apply the Chain Rule, set $u$ as $x^{2} - 3 x$ .

$\frac{d}{d u} [\cos (u)] \frac{d}{d x} [x^{2} - 3 x]$

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

$- \sin (u) \frac{d}{d x} [x^{2} - 3 x]$

Replace all occurrences of $u$ with $x^{2} - 3 x$ .

$- \sin (x^{2} - 3 x) \frac{d}{d x} [x^{2} - 3 x]$

Differentiate.

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By the Sum Rule, the derivative of $x^{2} - 3 x$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 3 x]$ .

$- \sin (x^{2} - 3 x) (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 3 x])$

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

$- \sin (x^{2} - 3 x) (2 x + \frac{d}{d x} [- 3 x])$

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

$- \sin (x^{2} - 3 x) (2 x - 3 \frac{d}{d x} [x])$

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

$- \sin (x^{2} - 3 x) (2 x - 3 \cdot 1)$

Multiply $- 3$ by $1$ .

$- \sin (x^{2} - 3 x) (2 x - 3)$

Find the Derivative – d/dx cos(x^2-3x)

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