Calculus Examples

$f (x) = 3 \cos^{2} (x) - 6 \sin (x)$

Step 1

This derivative could not be completed using the product rule. Mathway will use another method.

Step 2

By the Sum Rule, the derivative of $3 \cos^{2} (x) - 6 \sin (x)$ with respect to $x$ is $\frac{d}{d x} [3 \cos^{2} (x)] + \frac{d}{d x} [- 6 \sin (x)]$ .

$\frac{d}{d x} [3 \cos^{2} (x)] + \frac{d}{d x} [- 6 \sin (x)]$

Step 3

Evaluate $\frac{d}{d x} [3 \cos^{2} (x)]$ .

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Step 3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 \cos^{2} (x)$ with respect to $x$ is $3 \frac{d}{d x} [\cos^{2} (x)]$ .

$3 \frac{d}{d x} [\cos^{2} (x)] + \frac{d}{d x} [- 6 \sin (x)]$

Step 3.2

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

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Step 3.2.1

To apply the Chain Rule, set $u$ as $\cos (x)$ .

$3 (\frac{d}{d u} [u^{2}] \frac{d}{d x} [\cos (x)]) + \frac{d}{d x} [- 6 \sin (x)]$

Step 3.2.2

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

$3 (2 u \frac{d}{d x} [\cos (x)]) + \frac{d}{d x} [- 6 \sin (x)]$

Step 3.2.3

Replace all occurrences of $u$ with $\cos (x)$ .

$3 (2 \cos (x) \frac{d}{d x} [\cos (x)]) + \frac{d}{d x} [- 6 \sin (x)]$

Step 3.3

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

$3 (2 \cos (x) (- \sin (x))) + \frac{d}{d x} [- 6 \sin (x)]$

Step 3.4

Multiply $- 1$ by $2$ .

$3 (- 2 \cos (x) \sin (x)) + \frac{d}{d x} [- 6 \sin (x)]$

Step 3.5

Multiply $- 2$ by $3$ .

$- 6 \cos (x) \sin (x) + \frac{d}{d x} [- 6 \sin (x)]$

Step 4

Evaluate $\frac{d}{d x} [- 6 \sin (x)]$ .

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Step 4.1

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 \sin (x)$ with respect to $x$ is $- 6 \frac{d}{d x} [\sin (x)]$ .

$- 6 \cos (x) \sin (x) - 6 \frac{d}{d x} [\sin (x)]$

Step 4.2

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

$- 6 \cos (x) \sin (x) - 6 \cos (x)$