Calculus Examples

$f (x) = \frac{4}{3} - \frac{1}{2} x^{- 1} + \frac{1}{4} x^{4} + \frac{1}{2} x^{3}$

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 $\frac{4}{3} - \frac{1}{2} x^{- 1} + \frac{1}{4} x^{4} + \frac{1}{2} x^{3}$ with respect to $x$ is $\frac{d}{d x} [\frac{4}{3}] + \frac{d}{d x} [- \frac{1}{2} x^{- 1}] + \frac{d}{d x} [\frac{1}{4} x^{4}] + \frac{d}{d x} [\frac{1}{2} x^{3}]$ .

$\frac{d}{d x} [\frac{4}{3}] + \frac{d}{d x} [- \frac{1}{2} x^{- 1}] + \frac{d}{d x} [\frac{1}{4} x^{4}] + \frac{d}{d x} [\frac{1}{2} x^{3}]$

Step 1.1.2

Since $\frac{4}{3}$ is constant with respect to $x$ , the derivative of $\frac{4}{3}$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- \frac{1}{2} x^{- 1}] + \frac{d}{d x} [\frac{1}{4} x^{4}] + \frac{d}{d x} [\frac{1}{2} x^{3}]$

Step 1.2

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

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Step 1.2.1

Since $- \frac{1}{2}$ is constant with respect to $x$ , the derivative of $- \frac{1}{2} x^{- 1}$ with respect to $x$ is $- \frac{1}{2} \frac{d}{d x} [x^{- 1}]$ .

$0 - \frac{1}{2} \frac{d}{d x} [x^{- 1}] + \frac{d}{d x} [\frac{1}{4} x^{4}] + \frac{d}{d x} [\frac{1}{2} x^{3}]$

Step 1.2.2

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

$0 - \frac{1}{2} (- x^{- 2}) + \frac{d}{d x} [\frac{1}{4} x^{4}] + \frac{d}{d x} [\frac{1}{2} x^{3}]$

Step 1.2.3

Multiply $- 1$ by $- 1$ .

$0 + 1 (\frac{1}{2}) x^{- 2} + \frac{d}{d x} [\frac{1}{4} x^{4}] + \frac{d}{d x} [\frac{1}{2} x^{3}]$

Step 1.2.4

Multiply $\frac{1}{2}$ by $1$ .

$0 + \frac{1}{2} x^{- 2} + \frac{d}{d x} [\frac{1}{4} x^{4}] + \frac{d}{d x} [\frac{1}{2} x^{3}]$

Step 1.2.5

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

$0 + \frac{x^{- 2}}{2} + \frac{d}{d x} [\frac{1}{4} x^{4}] + \frac{d}{d x} [\frac{1}{2} x^{3}]$

Step 1.2.6

Move $x^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$0 + \frac{1}{2 x^{2}} + \frac{d}{d x} [\frac{1}{4} x^{4}] + \frac{d}{d x} [\frac{1}{2} x^{3}]$

Step 1.3

Evaluate $\frac{d}{d x} [\frac{1}{4} x^{4}]$ .

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Step 1.3.1

Since $\frac{1}{4}$ is constant with respect to $x$ , the derivative of $\frac{1}{4} x^{4}$ with respect to $x$ is $\frac{1}{4} \frac{d}{d x} [x^{4}]$ .

$0 + \frac{1}{2 x^{2}} + \frac{1}{4} \frac{d}{d x} [x^{4}] + \frac{d}{d x} [\frac{1}{2} x^{3}]$

Step 1.3.2

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

$0 + \frac{1}{2 x^{2}} + \frac{1}{4} (4 x^{3}) + \frac{d}{d x} [\frac{1}{2} x^{3}]$

Step 1.3.3

Combine $4$ and $\frac{1}{4}$ .

$0 + \frac{1}{2 x^{2}} + \frac{4}{4} x^{3} + \frac{d}{d x} [\frac{1}{2} x^{3}]$

Step 1.3.4

Combine $\frac{4}{4}$ and $x^{3}$ .

$0 + \frac{1}{2 x^{2}} + \frac{4 x^{3}}{4} + \frac{d}{d x} [\frac{1}{2} x^{3}]$

Step 1.3.5

Cancel the common factor of $4$ .

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Step 1.3.5.1

Cancel the common factor.

$0 + \frac{1}{2 x^{2}} + \frac{4 x^{3}}{4} + \frac{d}{d x} [\frac{1}{2} x^{3}]$

Step 1.3.5.2

Divide $x^{3}$ by $1$ .

$0 + \frac{1}{2 x^{2}} + x^{3} + \frac{d}{d x} [\frac{1}{2} x^{3}]$

Step 1.4

Evaluate $\frac{d}{d x} [\frac{1}{2} x^{3}]$ .

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Step 1.4.1

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{1}{2} x^{3}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [x^{3}]$ .

$0 + \frac{1}{2 x^{2}} + x^{3} + \frac{1}{2} \frac{d}{d x} [x^{3}]$

Step 1.4.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

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

Step 1.4.3

Combine $3$ and $\frac{1}{2}$ .

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

Step 1.4.4

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

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

Step 1.5

Simplify.

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Step 1.5.1

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

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

Step 1.5.2

Reorder terms.

$f' (x) = x^{3} + \frac{3 x^{2}}{2} + \frac{1}{2 x^{2}}$

Step 2

Find the second derivative.

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Step 2.1

Differentiate.

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Step 2.1.1

By the Sum Rule, the derivative of $x^{3} + \frac{3 x^{2}}{2} + \frac{1}{2 x^{2}}$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [\frac{3 x^{2}}{2}] + \frac{d}{d x} [\frac{1}{2 x^{2}}]$ .

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

Step 2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

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

Step 2.2

Evaluate $\frac{d}{d x} [\frac{3 x^{2}}{2}]$ .

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Step 2.2.1

Since $\frac{3}{2}$ is constant with respect to $x$ , the derivative of $\frac{3 x^{2}}{2}$ with respect to $x$ is $\frac{3}{2} \frac{d}{d x} [x^{2}]$ .

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

Step 2.2.2

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

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

Step 2.2.3

Combine $2$ and $\frac{3}{2}$ .

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

Step 2.2.4

Multiply $2$ by $3$ .

$3 x^{2} + \frac{6}{2} x + \frac{d}{d x} [\frac{1}{2 x^{2}}]$

Step 2.2.5

Combine $\frac{6}{2}$ and $x$ .

$3 x^{2} + \frac{6 x}{2} + \frac{d}{d x} [\frac{1}{2 x^{2}}]$

Step 2.2.6

Cancel the common factor of $6$ and $2$ .

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Step 2.2.6.1

Factor $2$ out of $6 x$ .

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

Step 2.2.6.2

Cancel the common factors.

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Step 2.2.6.2.1

Factor $2$ out of $2$ .

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

Step 2.2.6.2.2

Cancel the common factor.

$3 x^{2} + \frac{2 (3 x)}{2 \cdot 1} + \frac{d}{d x} [\frac{1}{2 x^{2}}]$

Step 2.2.6.2.3

Rewrite the expression.

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

Step 2.2.6.2.4

Divide $3 x$ by $1$ .

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

Step 2.3

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

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Step 2.3.1

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{1}{2 x^{2}}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [\frac{1}{x^{2}}]$ .

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

Step 2.3.2

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

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

Step 2.3.3

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 2.3.3.1

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

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

Step 2.3.3.2

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

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

Step 2.3.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

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

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Step 2.3.5.1

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

$3 x^{2} + 3 x + \frac{1}{2} (- x^{2 \cdot - 2} (2 x))$

Step 2.3.5.2

Multiply $2$ by $- 2$ .

$3 x^{2} + 3 x + \frac{1}{2} (- x^{- 4} (2 x))$

Step 2.3.6

Multiply $2$ by $- 1$ .

$3 x^{2} + 3 x + \frac{1}{2} (- 2 x^{- 4} x)$

Step 2.3.7

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

$3 x^{2} + 3 x + \frac{1}{2} (- 2 (x^{1} x^{- 4}))$

Step 2.3.8

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

$3 x^{2} + 3 x + \frac{1}{2} (- 2 x^{1 - 4})$

Step 2.3.9

Subtract $4$ from $1$ .

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

Step 2.3.10

Combine $- 2$ and $\frac{1}{2}$ .

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

Step 2.3.11

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

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

Step 2.3.12

Move $x^{- 3}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 2.3.13

Cancel the common factor of $- 2$ and $2$ .

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Step 2.3.13.1

Factor $2$ out of $- 2$ .

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

Step 2.3.13.2

Cancel the common factors.

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Step 2.3.13.2.1

Factor $2$ out of $2 x^{3}$ .

$3 x^{2} + 3 x + \frac{2 \cdot - 1}{2 (x^{3})}$

Step 2.3.13.2.2

Cancel the common factor.

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

Step 2.3.13.2.3

Rewrite the expression.

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

Step 2.3.14

Move the negative in front of the fraction.

$f'' (x) = 3 x^{2} + 3 x - \frac{1}{x^{3}}$

Step 3

Find the third derivative.

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Step 3.1

By the Sum Rule, the derivative of $3 x^{2} + 3 x - \frac{1}{x^{3}}$ with respect to $x$ is $\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [- \frac{1}{x^{3}}]$ .

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

Step 3.2

Evaluate $\frac{d}{d x} [3 x^{2}]$ .

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Step 3.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

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

Step 3.2.2

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

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

Step 3.2.3

Multiply $2$ by $3$ .

$6 x + \frac{d}{d x} [3 x] + \frac{d}{d x} [- \frac{1}{x^{3}}]$

Step 3.3

Evaluate $\frac{d}{d x} [3 x]$ .

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Step 3.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$6 x + 3 \frac{d}{d x} [x] + \frac{d}{d x} [- \frac{1}{x^{3}}]$

Step 3.3.2

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

$6 x + 3 \cdot 1 + \frac{d}{d x} [- \frac{1}{x^{3}}]$

Step 3.3.3

Multiply $3$ by $1$ .

$6 x + 3 + \frac{d}{d x} [- \frac{1}{x^{3}}]$

Step 3.4

Evaluate $\frac{d}{d x} [- \frac{1}{x^{3}}]$ .

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Step 3.4.1

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 $g (x) = \frac{1}{x^{3}}$ .

$6 x + 3 - \frac{d}{d x} [\frac{1}{x^{3}}] + \frac{1}{x^{3}} \frac{d}{d x} [- 1]$

Step 3.4.2

Rewrite $\frac{1}{x^{3}}$ as ${(x^{3})}^{- 1}$ .

$6 x + 3 - \frac{d}{d x} [{(x^{3})}^{- 1}] + \frac{1}{x^{3}} \frac{d}{d x} [- 1]$

Step 3.4.3

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 $g (x) = x^{3}$ .

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Step 3.4.3.1

To apply the Chain Rule, set $u$ as $x^{3}$ .

$6 x + 3 - (\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [x^{3}]) + \frac{1}{x^{3}} \frac{d}{d x} [- 1]$

Step 3.4.3.2

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

$6 x + 3 - (- u^{- 2} \frac{d}{d x} [x^{3}]) + \frac{1}{x^{3}} \frac{d}{d x} [- 1]$

Step 3.4.3.3

Replace all occurrences of $u$ with $x^{3}$ .

$6 x + 3 - (- {(x^{3})}^{- 2} \frac{d}{d x} [x^{3}]) + \frac{1}{x^{3}} \frac{d}{d x} [- 1]$

Step 3.4.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$6 x + 3 - (- {(x^{3})}^{- 2} (3 x^{2})) + \frac{1}{x^{3}} \frac{d}{d x} [- 1]$

Step 3.4.5

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

$6 x + 3 - (- {(x^{3})}^{- 2} (3 x^{2})) + \frac{1}{x^{3}} \cdot 0$

Step 3.4.6

Multiply the exponents in ${(x^{3})}^{- 2}$ .

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Step 3.4.6.1

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

$6 x + 3 - (- x^{3 \cdot - 2} (3 x^{2})) + \frac{1}{x^{3}} \cdot 0$

Step 3.4.6.2

Multiply $3$ by $- 2$ .

$6 x + 3 - (- x^{- 6} (3 x^{2})) + \frac{1}{x^{3}} \cdot 0$

Step 3.4.7

Multiply $3$ by $- 1$ .

$6 x + 3 - (- 3 x^{- 6} x^{2}) + \frac{1}{x^{3}} \cdot 0$

Step 3.4.8

Multiply $x^{- 6}$ by $x^{2}$ by adding the exponents.

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Step 3.4.8.1

Move $x^{2}$ .

$6 x + 3 - (- 3 (x^{2} x^{- 6})) + \frac{1}{x^{3}} \cdot 0$

Step 3.4.8.2

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

$6 x + 3 - (- 3 x^{2 - 6}) + \frac{1}{x^{3}} \cdot 0$

Step 3.4.8.3

Subtract $6$ from $2$ .

$6 x + 3 - (- 3 x^{- 4}) + \frac{1}{x^{3}} \cdot 0$

Step 3.4.9

Multiply $- 3$ by $- 1$ .

$6 x + 3 + 3 x^{- 4} + \frac{1}{x^{3}} \cdot 0$

Step 3.4.10

Multiply $\frac{1}{x^{3}}$ by $0$ .

$6 x + 3 + 3 x^{- 4} + 0$

Step 3.4.11

Add $3 x^{- 4}$ and $0$ .

$6 x + 3 + 3 x^{- 4}$

Step 3.5

Simplify.

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Step 3.5.1

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

$6 x + 3 + 3 \frac{1}{x^{4}}$

Step 3.5.2

Combine $3$ and $\frac{1}{x^{4}}$ .

$6 x + 3 + \frac{3}{x^{4}}$

Step 3.5.3

Reorder terms.

$f''' (x) = 6 x + \frac{3}{x^{4}} + 3$

Step 4

The third derivative of $f (x)$ with respect to $x$ is $6 x + \frac{3}{x^{4}} + 3$ .

$6 x + \frac{3}{x^{4}} + 3$