Precalculus Examples

$f (x) = 4 x e^{x} \csc (x)$

Step 1

Since $4$ is constant with respect to $x$ , the derivative of $4 x e^{x} \csc (x)$ with respect to $x$ is $4 \frac{d}{d x} [x e^{x} \csc (x)]$ .

$4 \frac{d}{d x} [x e^{x} \csc (x)]$

Step 2

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 $f (x) = x e^{x}$ and $g (x) = \csc (x)$ .

$4 (x e^{x} \frac{d}{d x} [\csc (x)] + \csc (x) \frac{d}{d x} [x e^{x}])$

Step 3

The derivative of $\csc (x)$ with respect to $x$ is $- \csc (x) \cot (x)$ .

$4 (x e^{x} (- \csc (x) \cot (x)) + \csc (x) \frac{d}{d x} [x e^{x}])$

Step 4

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 $g (x) = e^{x}$ .

$4 (x e^{x} (- \csc (x) \cot (x)) + \csc (x) (x \frac{d}{d x} [e^{x}] + e^{x} \frac{d}{d x} [x]))$

Step 5

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

$4 (x e^{x} (- \csc (x) \cot (x)) + \csc (x) (x e^{x} + e^{x} \frac{d}{d x} [x]))$

Step 6

Differentiate using the Power Rule.

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Step 6.1

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

$4 (x e^{x} (- \csc (x) \cot (x)) + \csc (x) (x e^{x} + e^{x} \cdot 1))$

Step 6.2

Multiply $e^{x}$ by $1$ .

$4 (x e^{x} (- \csc (x) \cot (x)) + \csc (x) (x e^{x} + e^{x}))$

Step 7

Simplify.

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Step 7.1

Apply the distributive property.

$4 (x e^{x} (- \csc (x) \cot (x)) + \csc (x) (x e^{x}) + \csc (x) e^{x})$

Step 7.2

Apply the distributive property.

$4 (x e^{x} (- \csc (x) \cot (x))) + 4 (\csc (x) (x e^{x})) + 4 (\csc (x) e^{x})$

Step 7.3

Multiply $- 1$ by $4$ .

$- 4 x e^{x} (\csc (x) \cot (x)) + 4 \csc (x) (x e^{x}) + 4 \csc (x) e^{x}$

Step 7.4

Reorder terms.

$- 4 e^{x} x \cot (x) \csc (x) + 4 e^{x} x \csc (x) + 4 e^{x} \csc (x)$