Calculus Examples

$f (x) = - \frac{1}{x}$

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 $f (x) = - 1$ and $g (x) = \frac{1}{x}$ .

$- \frac{d}{d x} [\frac{1}{x}] + \frac{1}{x} \frac{d}{d x} [- 1]$

Step 2

Differentiate.

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Step 2.1

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$- \frac{d}{d x} [x^{- 1}] + \frac{1}{x} \frac{d}{d x} [- 1]$

Step 2.2

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

$- - x^{- 2} + \frac{1}{x} \frac{d}{d x} [- 1]$

Step 2.3

Multiply.

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Step 2.3.1

Multiply $- 1$ by $- 1$ .

$1 x^{- 2} + \frac{1}{x} \frac{d}{d x} [- 1]$

Step 2.3.2

Multiply $x^{- 2}$ by $1$ .

$x^{- 2} + \frac{1}{x} \frac{d}{d x} [- 1]$

Step 2.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$x^{- 2} + \frac{1}{x} \cdot 0$

Step 2.5

Simplify the expression.

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Step 2.5.1

Multiply $\frac{1}{x}$ by $0$ .

$x^{- 2} + 0$

Step 2.5.2

Add $x^{- 2}$ and $0$ .

$x^{- 2}$

Step 3

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

$\frac{1}{x^{2}}$