Calculus Examples

$f (x) = (x^{3}) \sqrt[5]{2 - x}$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{2 - x}$ as ${(2 - x)}^{\frac{1}{5}}$ .

$\frac{d}{d x} [x^{3} {(2 - x)}^{\frac{1}{5}}]$

Step 2

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 $f (x) = x^{3}$ and $g (x) = {(2 - x)}^{\frac{1}{5}}$ .

$x^{3} \frac{d}{d x} [{(2 - x)}^{\frac{1}{5}}] + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{5}}$ and $g (x) = 2 - x$ .

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Step 3.1

To apply the Chain Rule, set $u$ as $2 - x$ .

$x^{3} (\frac{d}{d u} [u^{\frac{1}{5}}] \frac{d}{d x} [2 - x]) + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{1}{5}$ .

$x^{3} (\frac{1}{5} u^{\frac{1}{5} - 1} \frac{d}{d x} [2 - x]) + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 3.3

Replace all occurrences of $u$ with $2 - x$ .

$x^{3} (\frac{1}{5} {(2 - x)}^{\frac{1}{5} - 1} \frac{d}{d x} [2 - x]) + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 4

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$x^{3} (\frac{1}{5} {(2 - x)}^{\frac{1}{5} - 1 \cdot \frac{5}{5}} \frac{d}{d x} [2 - x]) + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 5

Combine $- 1$ and $\frac{5}{5}$ .

$x^{3} (\frac{1}{5} {(2 - x)}^{\frac{1}{5} + \frac{- 1 \cdot 5}{5}} \frac{d}{d x} [2 - x]) + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 6

Combine the numerators over the common denominator.

$x^{3} (\frac{1}{5} {(2 - x)}^{\frac{1 - 1 \cdot 5}{5}} \frac{d}{d x} [2 - x]) + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 7

Simplify the numerator.

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Step 7.1

Multiply $- 1$ by $5$ .

$x^{3} (\frac{1}{5} {(2 - x)}^{\frac{1 - 5}{5}} \frac{d}{d x} [2 - x]) + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 7.2

Subtract $5$ from $1$ .

$x^{3} (\frac{1}{5} {(2 - x)}^{\frac{- 4}{5}} \frac{d}{d x} [2 - x]) + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 8

Combine fractions.

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Step 8.1

Move the negative in front of the fraction.

$x^{3} (\frac{1}{5} {(2 - x)}^{- \frac{4}{5}} \frac{d}{d x} [2 - x]) + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 8.2

Combine $\frac{1}{5}$ and ${(2 - x)}^{- \frac{4}{5}}$ .

$x^{3} (\frac{{(2 - x)}^{- \frac{4}{5}}}{5} \frac{d}{d x} [2 - x]) + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 8.3

Move ${(2 - x)}^{- \frac{4}{5}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$x^{3} (\frac{1}{5 {(2 - x)}^{\frac{4}{5}}} \frac{d}{d x} [2 - x]) + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 8.4

Combine $\frac{1}{5 {(2 - x)}^{\frac{4}{5}}}$ and $x^{3}$ .

$\frac{x^{3}}{5 {(2 - x)}^{\frac{4}{5}}} \frac{d}{d x} [2 - x] + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 9

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

$\frac{x^{3}}{5 {(2 - x)}^{\frac{4}{5}}} (\frac{d}{d x} [2] + \frac{d}{d x} [- x]) + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 10

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$\frac{x^{3}}{5 {(2 - x)}^{\frac{4}{5}}} (0 + \frac{d}{d x} [- x]) + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 11

Add $0$ and $\frac{d}{d x} [- x]$ .

$\frac{x^{3}}{5 {(2 - x)}^{\frac{4}{5}}} \frac{d}{d x} [- x] + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 12

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

$\frac{x^{3}}{5 {(2 - x)}^{\frac{4}{5}}} (- \frac{d}{d x} [x]) + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 13

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

$\frac{x^{3}}{5 {(2 - x)}^{\frac{4}{5}}} (- 1 \cdot 1) + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 14

Combine fractions.

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Step 14.1

Multiply $- 1$ by $1$ .

$\frac{x^{3}}{5 {(2 - x)}^{\frac{4}{5}}} \cdot - 1 + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 14.2

Combine $\frac{x^{3}}{5 {(2 - x)}^{\frac{4}{5}}}$ and $- 1$ .

$\frac{x^{3} \cdot - 1}{5 {(2 - x)}^{\frac{4}{5}}} + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 14.3

Simplify the expression.

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Step 14.3.1

Move $- 1$ to the left of $x^{3}$ .

$\frac{- 1 \cdot x^{3}}{5 {(2 - x)}^{\frac{4}{5}}} + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 14.3.2

Rewrite $- 1 x^{3}$ as $- x^{3}$ .

$\frac{- x^{3}}{5 {(2 - x)}^{\frac{4}{5}}} + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 14.3.3

Move the negative in front of the fraction.

$- \frac{x^{3}}{5 {(2 - x)}^{\frac{4}{5}}} + {(2 - x)}^{\frac{1}{5}} \frac{d}{d x} [x^{3}]$

Step 15

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

$- \frac{x^{3}}{5 {(2 - x)}^{\frac{4}{5}}} + {(2 - x)}^{\frac{1}{5}} (3 x^{2})$

Step 16

Reorder.

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Step 16.1

Move $3$ to the left of ${(2 - x)}^{\frac{1}{5}}$ .

$- \frac{x^{3}}{5 {(2 - x)}^{\frac{4}{5}}} + 3 \cdot {(2 - x)}^{\frac{1}{5}} x^{2}$

Step 16.2

Move ${(- x + 2)}^{\frac{1}{5}}$ .

$- \frac{x^{3}}{5 {(- x + 2)}^{\frac{4}{5}}} + 3 x^{2} {(- x + 2)}^{\frac{1}{5}}$

Step 17

To write $3 x^{2} {(- x + 2)}^{\frac{1}{5}}$ as a fraction with a common denominator, multiply by $\frac{5 {(- x + 2)}^{\frac{4}{5}}}{5 {(- x + 2)}^{\frac{4}{5}}}$ .

$- \frac{x^{3}}{5 {(- x + 2)}^{\frac{4}{5}}} + 3 x^{2} {(- x + 2)}^{\frac{1}{5}} \cdot \frac{5 {(- x + 2)}^{\frac{4}{5}}}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 18

Combine $3 x^{2} {(- x + 2)}^{\frac{1}{5}}$ and $\frac{5 {(- x + 2)}^{\frac{4}{5}}}{5 {(- x + 2)}^{\frac{4}{5}}}$ .

$- \frac{x^{3}}{5 {(- x + 2)}^{\frac{4}{5}}} + \frac{3 x^{2} {(- x + 2)}^{\frac{1}{5}} (5 {(- x + 2)}^{\frac{4}{5}})}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 19

Combine the numerators over the common denominator.

$\frac{- x^{3} + 3 x^{2} {(- x + 2)}^{\frac{1}{5}} (5 {(- x + 2)}^{\frac{4}{5}})}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 20

Multiply $5$ by $3$ .

$\frac{- x^{3} + 15 x^{2} {(- x + 2)}^{\frac{1}{5}} {(- x + 2)}^{\frac{4}{5}}}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 21

Multiply ${(- x + 2)}^{\frac{1}{5}}$ by ${(- x + 2)}^{\frac{4}{5}}$ by adding the exponents.

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Step 21.1

Move ${(- x + 2)}^{\frac{4}{5}}$ .

$\frac{- x^{3} + 15 x^{2} ({(- x + 2)}^{\frac{4}{5}} {(- x + 2)}^{\frac{1}{5}})}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 21.2

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

$\frac{- x^{3} + 15 x^{2} {(- x + 2)}^{\frac{4}{5} + \frac{1}{5}}}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 21.3

Combine the numerators over the common denominator.

$\frac{- x^{3} + 15 x^{2} {(- x + 2)}^{\frac{4 + 1}{5}}}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 21.4

Add $4$ and $1$ .

$\frac{- x^{3} + 15 x^{2} {(- x + 2)}^{\frac{5}{5}}}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 21.5

Divide $5$ by $5$ .

$\frac{- x^{3} + 15 x^{2} {(- x + 2)}^{1}}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 22

Simplify $15 x^{2} {(- x + 2)}^{1}$ .

$\frac{- x^{3} + 15 x^{2} (- x + 2)}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23

Simplify.

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Step 23.1

Apply the distributive property.

$\frac{- x^{3} + 15 x^{2} (- x) + 15 x^{2} \cdot 2}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.2

Simplify the numerator.

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Step 23.2.1

Simplify each term.

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Step 23.2.1.1

Rewrite using the commutative property of multiplication.

$\frac{- x^{3} + 15 \cdot - 1 x^{2} x + 15 x^{2} \cdot 2}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.2.1.2

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

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Step 23.2.1.2.1

Move $x$ .

$\frac{- x^{3} + 15 \cdot - 1 (x \cdot x^{2}) + 15 x^{2} \cdot 2}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.2.1.2.2

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

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Step 23.2.1.2.2.1

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

$\frac{- x^{3} + 15 \cdot - 1 (x^{1} x^{2}) + 15 x^{2} \cdot 2}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.2.1.2.2.2

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

$\frac{- x^{3} + 15 \cdot - 1 x^{1 + 2} + 15 x^{2} \cdot 2}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.2.1.2.3

Add $1$ and $2$ .

$\frac{- x^{3} + 15 \cdot - 1 x^{3} + 15 x^{2} \cdot 2}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.2.1.3

Multiply $15$ by $- 1$ .

$\frac{- x^{3} - 15 x^{3} + 15 x^{2} \cdot 2}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.2.1.4

Multiply $2$ by $15$ .

$\frac{- x^{3} - 15 x^{3} + 30 x^{2}}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.2.2

Subtract $15 x^{3}$ from $- x^{3}$ .

$\frac{- 16 x^{3} + 30 x^{2}}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.3

Factor $2 x^{2}$ out of $- 16 x^{3} + 30 x^{2}$ .

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Step 23.3.1

Factor $2 x^{2}$ out of $- 16 x^{3}$ .

$\frac{2 x^{2} (- 8 x) + 30 x^{2}}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.3.2

Factor $2 x^{2}$ out of $30 x^{2}$ .

$\frac{2 x^{2} (- 8 x) + 2 x^{2} (15)}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.3.3

Factor $2 x^{2}$ out of $2 x^{2} (- 8 x) + 2 x^{2} (15)$ .

$\frac{2 x^{2} (- 8 x + 15)}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.4

Factor $- 1$ out of $- 8 x$ .

$\frac{2 x^{2} (- (8 x) + 15)}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.5

Rewrite $15$ as $- 1 (- 15)$ .

$\frac{2 x^{2} (- (8 x) - 1 (- 15))}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.6

Factor $- 1$ out of $- (8 x) - 1 (- 15)$ .

$\frac{2 x^{2} (- (8 x - 15))}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.7

Rewrite $- (8 x - 15)$ as $- 1 (8 x - 15)$ .

$\frac{2 x^{2} (- 1 (8 x - 15))}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.8

Move the negative in front of the fraction.

$- \frac{(2 x^{2}) (8 x - 15)}{5 {(- x + 2)}^{\frac{4}{5}}}$

Step 23.9

Reorder factors in $- \frac{(2 x^{2}) (8 x - 15)}{5 {(- x + 2)}^{\frac{4}{5}}}$ .

$- \frac{2 x^{2} (8 x - 15)}{5 {(- x + 2)}^{\frac{4}{5}}}$