Calculus Examples

$x \cos (2 x)$

Step 1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = \cos (2 x)$ .

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

Step 2

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

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Step 2.1

To apply the Chain Rule, set $u$ as $2 x$ .

$x (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [2 x]) + \cos (2 x) \frac{d}{d x} [x]$

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u$ with $2 x$ .

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

Step 3

Differentiate.

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Step 3.1

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

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

Step 3.2

Multiply $2$ by $- 1$ .

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

Step 3.3

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

$x (- 2 \sin (2 x) \cdot 1) + \cos (2 x) \frac{d}{d x} [x]$

Step 3.4

Multiply $- 2$ by $1$ .

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

Step 3.5

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

$x (- 2 \sin (2 x)) + \cos (2 x) \cdot 1$

Step 3.6

Simplify the expression.

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Step 3.6.1

Multiply $\cos (2 x)$ by $1$ .

$x (- 2 \sin (2 x)) + \cos (2 x)$

Step 3.6.2

Reorder terms.

$- 2 x \sin (2 x) + \cos (2 x)$