Calculus Examples

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

Step 1

Find the first derivative.

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Step 1.1

Differentiate.

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Step 1.1.1

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

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

Step 1.1.2

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

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.2.4

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

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

Step 1.2.5

Multiply $- 1$ by $- 1$ .

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

Step 1.2.6

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

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

Step 1.2.7

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

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

Step 1.2.8

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

$1 + x^{- 2}$

Step 1.3

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

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

Step 1.4

Reorder terms.

$f' (x) = \frac{1}{x^{2}} + 1$

Step 2

Find the second derivative.

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Step 2.1

By the Sum Rule, the derivative of $\frac{1}{x^{2}} + 1$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{x^{2}}] + \frac{d}{d x} [1]$ .

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

Step 2.2

Evaluate $\frac{d}{d x} [\frac{1}{x^{2}}]$ .

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Step 2.2.1

Rewrite $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

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

Step 2.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{2}$ .

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Step 2.2.2.1

To apply the Chain Rule, set $u$ as $x^{2}$ .

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

Step 2.2.2.2

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

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

Step 2.2.2.3

Replace all occurrences of $u$ with $x^{2}$ .

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

Step 2.2.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 2.2.4

Multiply the exponents in ${(x^{2})}^{- 2}$ .

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Step 2.2.4.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 2.2.4.2

Multiply $2$ by $- 2$ .

$- x^{- 4} (2 x) + \frac{d}{d x} [1]$

Step 2.2.5

Multiply $2$ by $- 1$ .

$- 2 x^{- 4} x + \frac{d}{d x} [1]$

Step 2.2.6

Raise $x$ to the power of $1$ .

$- 2 (x^{1} x^{- 4}) + \frac{d}{d x} [1]$

Step 2.2.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- 2 x^{1 - 4} + \frac{d}{d x} [1]$

Step 2.2.8

Subtract $4$ from $1$ .

$- 2 x^{- 3} + \frac{d}{d x} [1]$

Step 2.3

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

$- 2 x^{- 3} + 0$

Step 2.4

Simplify.

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Step 2.4.1

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

$- 2 \frac{1}{x^{3}} + 0$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Combine $- 2$ and $\frac{1}{x^{3}}$ .

$\frac{- 2}{x^{3}} + 0$

Step 2.4.2.2

Move the negative in front of the fraction.

$- \frac{2}{x^{3}} + 0$

Step 2.4.2.3

Add $- \frac{2}{x^{3}}$ and $0$ .

$f'' (x) = - \frac{2}{x^{3}}$