Calculus Examples

$y = e^{2 x \cos (x)}$

Step 1

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

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Step 1.1

To apply the Chain Rule, set $u$ as $2 x \cos (x)$ .

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

Step 1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

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

Step 1.3

Replace all occurrences of $u$ with $2 x \cos (x)$ .

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

Step 2

Differentiate using the Constant Multiple Rule.

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Step 2.1

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

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

Step 2.2

Move $2$ to the left of $e^{2 x \cos (x)}$ .

$2 \cdot e^{2 x \cos (x)} \frac{d}{d x} [x \cos (x)]$

Step 3

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 $g (x) = \cos (x)$ .

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

Step 4

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

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

Step 5

Differentiate using the Power Rule.

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Step 5.1

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

$2 e^{2 x \cos (x)} (x (- \sin (x)) + \cos (x) \cdot 1)$

Step 5.2

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

$2 e^{2 x \cos (x)} (x (- \sin (x)) + \cos (x))$

Step 6

Simplify.

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Step 6.1

Apply the distributive property.

$2 e^{2 x \cos (x)} (x (- \sin (x))) + 2 e^{2 x \cos (x)} \cos (x)$

Step 6.2

Multiply $- 1$ by $2$ .

$- 2 e^{2 x \cos (x)} x \sin (x) + 2 e^{2 x \cos (x)} \cos (x)$