Calculus Examples

$- 4 x^{- 4} + \sec (x) - e^{3}$

Step 1

By the Sum Rule, the derivative of $- 4 x^{- 4} + \sec (x) - e^{3}$ with respect to $x$ is $\frac{d}{d x} [- 4 x^{- 4}] + \frac{d}{d x} [\sec (x)] + \frac{d}{d x} [- e^{3}]$ .

$\frac{d}{d x} [- 4 x^{- 4}] + \frac{d}{d x} [\sec (x)] + \frac{d}{d x} [- e^{3}]$

Step 2

Evaluate $\frac{d}{d x} [- 4 x^{- 4}]$ .

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Step 2.1

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

$- 4 \frac{d}{d x} [x^{- 4}] + \frac{d}{d x} [\sec (x)] + \frac{d}{d x} [- e^{3}]$

Step 2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = - 4$ .

$- 4 (- 4 x^{- 5}) + \frac{d}{d x} [\sec (x)] + \frac{d}{d x} [- e^{3}]$

Step 2.3

Multiply $- 4$ by $- 4$ .

$16 x^{- 5} + \frac{d}{d x} [\sec (x)] + \frac{d}{d x} [- e^{3}]$

Step 3

The derivative of $\sec (x)$ with respect to $x$ is $\sec (x) \tan (x)$ .

$16 x^{- 5} + \sec (x) \tan (x) + \frac{d}{d x} [- e^{3}]$

Step 4

Since $- e^{3}$ is constant with respect to $x$ , the derivative of $- e^{3}$ with respect to $x$ is $0$ .

$16 x^{- 5} + \sec (x) \tan (x) + 0$

Step 5

Simplify.

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Step 5.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$16 \frac{1}{x^{5}} + \sec (x) \tan (x) + 0$

Step 5.2

Combine terms.

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Step 5.2.1

Combine $16$ and $\frac{1}{x^{5}}$ .

$\frac{16}{x^{5}} + \sec (x) \tan (x) + 0$

Step 5.2.2

Add $\frac{16}{x^{5}} + \sec (x) \tan (x)$ and $0$ .

$\frac{16}{x^{5}} + \sec (x) \tan (x)$

Step 5.3

Reorder terms.

$\sec (x) \tan (x) + \frac{16}{x^{5}}$