Calculus Examples

$- \frac{1}{x} + 1$

Step 1

By the Sum Rule, the derivative of $- \frac{1}{x} + 1$ with respect to $x$ is $\frac{d}{d x} [- \frac{1}{x}] + \frac{d}{d x} [1]$ .

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

Step 2

Evaluate $\frac{d}{d x} [- \frac{1}{x}]$ .

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Multiply $- 1$ by $- 1$ .

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

Step 2.6

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

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

Step 2.7

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

$x^{- 2} + 0 + \frac{d}{d x} [1]$

Step 2.8

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

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

Step 3

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

$x^{- 2} + 0$

Step 4

Simplify.

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Step 4.1

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

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

Step 4.2

Add $\frac{1}{x^{2}}$ and $0$ .

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