Calculus Examples

\frac{6}{x} - \frac{2}{x^{3}} + \frac{1}{x^{4}}

$\frac{6}{x} - \frac{2}{x^{3}} + \frac{1}{x^{4}}$

Step 1

By the Sum Rule, the derivative of

\frac{6}{x} - \frac{2}{x^{3}} + \frac{1}{x^{4}}

$\frac{6}{x} - \frac{2}{x^{3}} + \frac{1}{x^{4}}$ with respect to

x

$x$ is

\frac{d}{d x} [\frac{6}{x}] + \frac{d}{d x} [- \frac{2}{x^{3}}] + \frac{d}{d x} [\frac{1}{x^{4}}]

$\frac{d}{d x} [\frac{6}{x}] + \frac{d}{d x} [- \frac{2}{x^{3}}] + \frac{d}{d x} [\frac{1}{x^{4}}]$ .

\frac{d}{d x} [\frac{6}{x}] + \frac{d}{d x} [- \frac{2}{x^{3}}] + \frac{d}{d x} [\frac{1}{x^{4}}]

$\frac{d}{d x} [\frac{6}{x}] + \frac{d}{d x} [- \frac{2}{x^{3}}] + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 2

Evaluate

\frac{d}{d x} [\frac{6}{x}]

$\frac{d}{d x} [\frac{6}{x}]$ .

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Step 2.1

Since

6

$6$ is constant with respect to

x

$x$ , the derivative of

\frac{6}{x}

$\frac{6}{x}$ with respect to

x

$x$ is

6 \frac{d}{d x} [\frac{1}{x}]

$6 \frac{d}{d x} [\frac{1}{x}]$ .

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

$6 \frac{d}{d x} [\frac{1}{x}] + \frac{d}{d x} [- \frac{2}{x^{3}}] + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 2.2

Rewrite

\frac{1}{x}

$\frac{1}{x}$ as

x^{- 1}

$x^{- 1}$ .

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

$6 \frac{d}{d x} [x^{- 1}] + \frac{d}{d x} [- \frac{2}{x^{3}}] + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 2.3

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = - 1

$n = - 1$ .

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

$6 (- x^{- 2}) + \frac{d}{d x} [- \frac{2}{x^{3}}] + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 2.4

Multiply

- 1

$- 1$ by

6

$6$ .

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

$- 6 x^{- 2} + \frac{d}{d x} [- \frac{2}{x^{3}}] + \frac{d}{d x} [\frac{1}{x^{4}}]$

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

$- 6 x^{- 2} + \frac{d}{d x} [- \frac{2}{x^{3}}] + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 3

Evaluate

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

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

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Step 3.1

Since

- 2

$- 2$ is constant with respect to

x

$x$ , the derivative of

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

$- \frac{2}{x^{3}}$ with respect to

x

$x$ is

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

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

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

$- 6 x^{- 2} - 2 \frac{d}{d x} [\frac{1}{x^{3}}] + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 3.2

Rewrite

\frac{1}{x^{3}}

$\frac{1}{x^{3}}$ as

{(x^{3})}^{- 1}

${(x^{3})}^{- 1}$ .

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

$- 6 x^{- 2} - 2 \frac{d}{d x} [{(x^{3})}^{- 1}] + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 3.3

Differentiate using the chain rule, which states that

\frac{d}{d x} [f (g (x))]

$\frac{d}{d x} [f (g (x))]$ is

f' (g (x)) g' (x)

$f' (g (x)) g' (x)$ where

f (x) = x^{- 1}

$f (x) = x^{- 1}$ and

g (x) = x^{3}

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

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Step 3.3.1

To apply the Chain Rule, set

u_{1}

$u_{1}$ as

x^{3}

$x^{3}$ .

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

$- 6 x^{- 2} - 2 (\frac{d}{d u_{1}} [{u_{1}}^{- 1}] \frac{d}{d x} [x^{3}]) + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 3.3.2

Differentiate using the Power Rule which states that

\frac{d}{d u_{1}} [{u_{1}}^{n}]

$\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is

n {u_{1}}^{n - 1}

$n {u_{1}}^{n - 1}$ where

n = - 1

$n = - 1$ .

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

$- 6 x^{- 2} - 2 (- {u_{1}}^{- 2} \frac{d}{d x} [x^{3}]) + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 3.3.3

Replace all occurrences of

u_{1}

$u_{1}$ with

x^{3}

$x^{3}$ .

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

$- 6 x^{- 2} - 2 (- {(x^{3})}^{- 2} \frac{d}{d x} [x^{3}]) + \frac{d}{d x} [\frac{1}{x^{4}}]$

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

$- 6 x^{- 2} - 2 (- {(x^{3})}^{- 2} \frac{d}{d x} [x^{3}]) + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 3.4

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 3

$n = 3$ .

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

$- 6 x^{- 2} - 2 (- {(x^{3})}^{- 2} (3 x^{2})) + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 3.5

Multiply the exponents in

{(x^{3})}^{- 2}

${(x^{3})}^{- 2}$ .

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Step 3.5.1

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

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

$- 6 x^{- 2} - 2 (- x^{3 \cdot - 2} (3 x^{2})) + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 3.5.2

Multiply

3

$3$ by

- 2

$- 2$ .

- 6 x^{- 2} - 2 (- x^{- 6} (3 x^{2})) + \frac{d}{d x} [\frac{1}{x^{4}}]

$- 6 x^{- 2} - 2 (- x^{- 6} (3 x^{2})) + \frac{d}{d x} [\frac{1}{x^{4}}]$

- 6 x^{- 2} - 2 (- x^{- 6} (3 x^{2})) + \frac{d}{d x} [\frac{1}{x^{4}}]

$- 6 x^{- 2} - 2 (- x^{- 6} (3 x^{2})) + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 3.6

Multiply

3

$3$ by

- 1

$- 1$ .

- 6 x^{- 2} - 2 (- 3 x^{- 6} x^{2}) + \frac{d}{d x} [\frac{1}{x^{4}}]

$- 6 x^{- 2} - 2 (- 3 x^{- 6} x^{2}) + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 3.7

Multiply

x^{- 6}

$x^{- 6}$ by

x^{2}

$x^{2}$ by adding the exponents.

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Step 3.7.1

Move

x^{2}

$x^{2}$ .

- 6 x^{- 2} - 2 (- 3 (x^{2} x^{- 6})) + \frac{d}{d x} [\frac{1}{x^{4}}]

$- 6 x^{- 2} - 2 (- 3 (x^{2} x^{- 6})) + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 3.7.2

Use the power rule

a^{m} a^{n} = a^{m + n}

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

- 6 x^{- 2} - 2 (- 3 x^{2 - 6}) + \frac{d}{d x} [\frac{1}{x^{4}}]

$- 6 x^{- 2} - 2 (- 3 x^{2 - 6}) + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 3.7.3

Subtract

6

$6$ from

2

$2$ .

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

$- 6 x^{- 2} - 2 (- 3 x^{- 4}) + \frac{d}{d x} [\frac{1}{x^{4}}]$

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

$- 6 x^{- 2} - 2 (- 3 x^{- 4}) + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 3.8

Multiply

- 3

$- 3$ by

- 2

$- 2$ .

- 6 x^{- 2} + 6 x^{- 4} + \frac{d}{d x} [\frac{1}{x^{4}}]

$- 6 x^{- 2} + 6 x^{- 4} + \frac{d}{d x} [\frac{1}{x^{4}}]$

- 6 x^{- 2} + 6 x^{- 4} + \frac{d}{d x} [\frac{1}{x^{4}}]

$- 6 x^{- 2} + 6 x^{- 4} + \frac{d}{d x} [\frac{1}{x^{4}}]$

Step 4

Evaluate

\frac{d}{d x} [\frac{1}{x^{4}}]

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

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Step 4.1

Rewrite

\frac{1}{x^{4}}

$\frac{1}{x^{4}}$ as

{(x^{4})}^{- 1}

${(x^{4})}^{- 1}$ .

- 6 x^{- 2} + 6 x^{- 4} + \frac{d}{d x} [{(x^{4})}^{- 1}]

$- 6 x^{- 2} + 6 x^{- 4} + \frac{d}{d x} [{(x^{4})}^{- 1}]$

Step 4.2

Differentiate using the chain rule, which states that

\frac{d}{d x} [f (g (x))]

$\frac{d}{d x} [f (g (x))]$ is

f' (g (x)) g' (x)

$f' (g (x)) g' (x)$ where

f (x) = x^{- 1}

$f (x) = x^{- 1}$ and

g (x) = x^{4}

$g (x) = x^{4}$ .

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Step 4.2.1

To apply the Chain Rule, set

u_{2}

$u_{2}$ as

x^{4}

$x^{4}$ .

- 6 x^{- 2} + 6 x^{- 4} + \frac{d}{d u_{2}} [{u_{2}}^{- 1}] \frac{d}{d x} [x^{4}]

$- 6 x^{- 2} + 6 x^{- 4} + \frac{d}{d u_{2}} [{u_{2}}^{- 1}] \frac{d}{d x} [x^{4}]$

Step 4.2.2

Differentiate using the Power Rule which states that

\frac{d}{d u_{2}} [{u_{2}}^{n}]

$\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is

n {u_{2}}^{n - 1}

$n {u_{2}}^{n - 1}$ where

n = - 1

$n = - 1$ .

- 6 x^{- 2} + 6 x^{- 4} - {u_{2}}^{- 2} \frac{d}{d x} [x^{4}]

$- 6 x^{- 2} + 6 x^{- 4} - {u_{2}}^{- 2} \frac{d}{d x} [x^{4}]$

Step 4.2.3

Replace all occurrences of

u_{2}

$u_{2}$ with

x^{4}

$x^{4}$ .

- 6 x^{- 2} + 6 x^{- 4} - {(x^{4})}^{- 2} \frac{d}{d x} [x^{4}]

$- 6 x^{- 2} + 6 x^{- 4} - {(x^{4})}^{- 2} \frac{d}{d x} [x^{4}]$

- 6 x^{- 2} + 6 x^{- 4} - {(x^{4})}^{- 2} \frac{d}{d x} [x^{4}]

$- 6 x^{- 2} + 6 x^{- 4} - {(x^{4})}^{- 2} \frac{d}{d x} [x^{4}]$

Step 4.3

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 4

$n = 4$ .

- 6 x^{- 2} + 6 x^{- 4} - {(x^{4})}^{- 2} (4 x^{3})

$- 6 x^{- 2} + 6 x^{- 4} - {(x^{4})}^{- 2} (4 x^{3})$

Step 4.4

Multiply the exponents in

{(x^{4})}^{- 2}

${(x^{4})}^{- 2}$ .

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Step 4.4.1

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

- 6 x^{- 2} + 6 x^{- 4} - x^{4 \cdot - 2} (4 x^{3})

$- 6 x^{- 2} + 6 x^{- 4} - x^{4 \cdot - 2} (4 x^{3})$

Step 4.4.2

Multiply

4

$4$ by

- 2

$- 2$ .

- 6 x^{- 2} + 6 x^{- 4} - x^{- 8} (4 x^{3})

$- 6 x^{- 2} + 6 x^{- 4} - x^{- 8} (4 x^{3})$

- 6 x^{- 2} + 6 x^{- 4} - x^{- 8} (4 x^{3})

$- 6 x^{- 2} + 6 x^{- 4} - x^{- 8} (4 x^{3})$

Step 4.5

Multiply

4

$4$ by

- 1

$- 1$ .

- 6 x^{- 2} + 6 x^{- 4} - 4 x^{- 8} x^{3}

$- 6 x^{- 2} + 6 x^{- 4} - 4 x^{- 8} x^{3}$

Step 4.6

Multiply

x^{- 8}

$x^{- 8}$ by

x^{3}

$x^{3}$ by adding the exponents.

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Step 4.6.1

Move

x^{3}

$x^{3}$ .

- 6 x^{- 2} + 6 x^{- 4} - 4 (x^{3} x^{- 8})

$- 6 x^{- 2} + 6 x^{- 4} - 4 (x^{3} x^{- 8})$

Step 4.6.2

Use the power rule

a^{m} a^{n} = a^{m + n}

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

- 6 x^{- 2} + 6 x^{- 4} - 4 x^{3 - 8}

$- 6 x^{- 2} + 6 x^{- 4} - 4 x^{3 - 8}$

Step 4.6.3

Subtract

8

$8$ from

3

$3$ .

- 6 x^{- 2} + 6 x^{- 4} - 4 x^{- 5}

$- 6 x^{- 2} + 6 x^{- 4} - 4 x^{- 5}$

- 6 x^{- 2} + 6 x^{- 4} - 4 x^{- 5}

$- 6 x^{- 2} + 6 x^{- 4} - 4 x^{- 5}$

- 6 x^{- 2} + 6 x^{- 4} - 4 x^{- 5}

$- 6 x^{- 2} + 6 x^{- 4} - 4 x^{- 5}$

Step 5

Simplify.

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Step 5.1

Rewrite the expression using the negative exponent rule

b^{- n} = \frac{1}{b^{n}}

$b^{- n} = \frac{1}{b^{n}}$ .

- 6 \frac{1}{x^{2}} + 6 x^{- 4} - 4 x^{- 5}

$- 6 \frac{1}{x^{2}} + 6 x^{- 4} - 4 x^{- 5}$

Step 5.2

Rewrite the expression using the negative exponent rule

b^{- n} = \frac{1}{b^{n}}

$b^{- n} = \frac{1}{b^{n}}$ .

- 6 \frac{1}{x^{2}} + 6 \frac{1}{x^{4}} - 4 x^{- 5}

$- 6 \frac{1}{x^{2}} + 6 \frac{1}{x^{4}} - 4 x^{- 5}$

Step 5.3

Rewrite the expression using the negative exponent rule

b^{- n} = \frac{1}{b^{n}}

$b^{- n} = \frac{1}{b^{n}}$ .

- 6 \frac{1}{x^{2}} + 6 \frac{1}{x^{4}} - 4 \frac{1}{x^{5}}

$- 6 \frac{1}{x^{2}} + 6 \frac{1}{x^{4}} - 4 \frac{1}{x^{5}}$

Step 5.4

Combine terms.

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Step 5.4.1

Combine

- 6

$- 6$ and

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

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

\frac{- 6}{x^{2}} + 6 \frac{1}{x^{4}} - 4 \frac{1}{x^{5}}

$\frac{- 6}{x^{2}} + 6 \frac{1}{x^{4}} - 4 \frac{1}{x^{5}}$

Step 5.4.2

Move the negative in front of the fraction.

- \frac{6}{x^{2}} + 6 \frac{1}{x^{4}} - 4 \frac{1}{x^{5}}

$- \frac{6}{x^{2}} + 6 \frac{1}{x^{4}} - 4 \frac{1}{x^{5}}$

Step 5.4.3

Combine

6

$6$ and

\frac{1}{x^{4}}

$\frac{1}{x^{4}}$ .

- \frac{6}{x^{2}} + \frac{6}{x^{4}} - 4 \frac{1}{x^{5}}

$- \frac{6}{x^{2}} + \frac{6}{x^{4}} - 4 \frac{1}{x^{5}}$

Step 5.4.4

Combine

- 4

$- 4$ and

\frac{1}{x^{5}}

$\frac{1}{x^{5}}$ .

- \frac{6}{x^{2}} + \frac{6}{x^{4}} + \frac{- 4}{x^{5}}

$- \frac{6}{x^{2}} + \frac{6}{x^{4}} + \frac{- 4}{x^{5}}$

Step 5.4.5

Move the negative in front of the fraction.

- \frac{6}{x^{2}} + \frac{6}{x^{4}} - \frac{4}{x^{5}}

$- \frac{6}{x^{2}} + \frac{6}{x^{4}} - \frac{4}{x^{5}}$

- \frac{6}{x^{2}} + \frac{6}{x^{4}} - \frac{4}{x^{5}}

$- \frac{6}{x^{2}} + \frac{6}{x^{4}} - \frac{4}{x^{5}}$

- \frac{6}{x^{2}} + \frac{6}{x^{4}} - \frac{4}{x^{5}}

$- \frac{6}{x^{2}} + \frac{6}{x^{4}} - \frac{4}{x^{5}}$