Calculus Examples

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

Step 1

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

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

Step 2

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

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Step 2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

Step 2.2

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

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

Step 2.3

Multiply $- 1$ by $1$ .

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 4

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

$- 1 - x^{- 2} + 0$

Step 5

Simplify.

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Reorder terms.

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