Calculus Examples

$y = \sqrt[3]{8 - x^{3}}$

Step 1

Write $y = \sqrt[3]{8 - x^{3}}$ as a function.

$f (x) = \sqrt[3]{8 - x^{3}}$

Step 2

Find the first derivative of the function.

Tap for more steps...

Step 2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{8 - x^{3}}$ as ${(8 - x^{3})}^{\frac{1}{3}}$ .

$\frac{d}{d x} [{(8 - x^{3})}^{\frac{1}{3}}]$

Step 2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{3}}$ and $g (x) = 8 - x^{3}$ .

Tap for more steps...

Step 2.2.1

To apply the Chain Rule, set $u$ as $8 - x^{3}$ .

$\frac{d}{d u} [u^{\frac{1}{3}}] \frac{d}{d x} [8 - x^{3}]$

Step 2.2.2

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

$\frac{1}{3} u^{\frac{1}{3} - 1} \frac{d}{d x} [8 - x^{3}]$

Step 2.2.3

Replace all occurrences of $u$ with $8 - x^{3}$ .

$\frac{1}{3} {(8 - x^{3})}^{\frac{1}{3} - 1} \frac{d}{d x} [8 - x^{3}]$

Step 2.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{1}{3} {(8 - x^{3})}^{\frac{1}{3} - 1 \cdot \frac{3}{3}} \frac{d}{d x} [8 - x^{3}]$

Step 2.4

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

$\frac{1}{3} {(8 - x^{3})}^{\frac{1}{3} + \frac{- 1 \cdot 3}{3}} \frac{d}{d x} [8 - x^{3}]$

Step 2.5

Combine the numerators over the common denominator.

$\frac{1}{3} {(8 - x^{3})}^{\frac{1 - 1 \cdot 3}{3}} \frac{d}{d x} [8 - x^{3}]$

Step 2.6

Simplify the numerator.

Tap for more steps...

Step 2.6.1

Multiply $- 1$ by $3$ .

$\frac{1}{3} {(8 - x^{3})}^{\frac{1 - 3}{3}} \frac{d}{d x} [8 - x^{3}]$

Step 2.6.2

Subtract $3$ from $1$ .

$\frac{1}{3} {(8 - x^{3})}^{\frac{- 2}{3}} \frac{d}{d x} [8 - x^{3}]$

Step 2.7

Combine fractions.

Tap for more steps...

Step 2.7.1

Move the negative in front of the fraction.

$\frac{1}{3} {(8 - x^{3})}^{- \frac{2}{3}} \frac{d}{d x} [8 - x^{3}]$

Step 2.7.2

Combine $\frac{1}{3}$ and ${(8 - x^{3})}^{- \frac{2}{3}}$ .

$\frac{{(8 - x^{3})}^{- \frac{2}{3}}}{3} \frac{d}{d x} [8 - x^{3}]$

Step 2.7.3

Move ${(8 - x^{3})}^{- \frac{2}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}} \frac{d}{d x} [8 - x^{3}]$

Step 2.8

By the Sum Rule, the derivative of $8 - x^{3}$ with respect to $x$ is $\frac{d}{d x} [8] + \frac{d}{d x} [- x^{3}]$ .

$\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}} (\frac{d}{d x} [8] + \frac{d}{d x} [- x^{3}])$

Step 2.9

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}} (0 + \frac{d}{d x} [- x^{3}])$

Step 2.10

Add $0$ and $\frac{d}{d x} [- x^{3}]$ .

$\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}} \frac{d}{d x} [- x^{3}]$

Step 2.11

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

$\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}} (- \frac{d}{d x} [x^{3}])$

Step 2.12

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

$\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}} (- (3 x^{2}))$

Step 2.13

Simplify terms.

Tap for more steps...

Step 2.13.1

Multiply $3$ by $- 1$ .

$\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}} (- 3 x^{2})$

Step 2.13.2

Combine $- 3$ and $\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}}$ .

$\frac{- 3}{3 {(8 - x^{3})}^{\frac{2}{3}}} x^{2}$

Step 2.13.3

Combine $\frac{- 3}{3 {(8 - x^{3})}^{\frac{2}{3}}}$ and $x^{2}$ .

$\frac{- 3 x^{2}}{3 {(8 - x^{3})}^{\frac{2}{3}}}$

Step 2.13.4

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

$\frac{3 (- x^{2})}{3 {(8 - x^{3})}^{\frac{2}{3}}}$

Step 2.14

Cancel the common factors.

Tap for more steps...

Step 2.14.1

Factor $3$ out of $3 {(8 - x^{3})}^{\frac{2}{3}}$ .

$\frac{3 (- x^{2})}{3 ({(8 - x^{3})}^{\frac{2}{3}})}$

Step 2.14.2

Cancel the common factor.

$\frac{3 (- x^{2})}{3 {(8 - x^{3})}^{\frac{2}{3}}}$

Step 2.14.3

Rewrite the expression.

$\frac{- x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}}$

Step 2.15

Move the negative in front of the fraction.

$- \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}}$

Step 3

Find the second derivative of the function.

Tap for more steps...

Step 3.1

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 $g (x) = \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}}$ .

$f'' (x) = - \frac{d}{d x} \cdot \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.2

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x^{2}$ and $g (x) = {(8 - x^{3})}^{\frac{2}{3}}$ .

$f'' (x) = - \frac{{(8 - x^{3})}^{\frac{2}{3}} \frac{d}{d x} (x^{2}) - x^{2} \frac{d}{d x} {(8 - x^{3})}^{\frac{2}{3}}}{{({(8 - x^{3})}^{\frac{2}{3}})}^{2}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.3

Differentiate using the Power Rule.

Tap for more steps...

Step 3.3.1

Multiply the exponents in ${({(8 - x^{3})}^{\frac{2}{3}})}^{2}$ .

Tap for more steps...

Step 3.3.1.1

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

$f'' (x) = - \frac{{(8 - x^{3})}^{\frac{2}{3}} \frac{d}{d x} (x^{2}) - x^{2} \frac{d}{d x} {(8 - x^{3})}^{\frac{2}{3}}}{{(8 - x^{3})}^{\frac{2}{3} \cdot 2}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.3.1.2

Multiply $\frac{2}{3} \cdot 2$ .

Tap for more steps...

Step 3.3.1.2.1

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

$f'' (x) = - \frac{{(8 - x^{3})}^{\frac{2}{3}} \frac{d}{d x} (x^{2}) - x^{2} \frac{d}{d x} {(8 - x^{3})}^{\frac{2}{3}}}{{(8 - x^{3})}^{\frac{2 \cdot 2}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.3.1.2.2

Multiply $2$ by $2$ .

$f'' (x) = - \frac{{(8 - x^{3})}^{\frac{2}{3}} \frac{d}{d x} (x^{2}) - x^{2} \frac{d}{d x} {(8 - x^{3})}^{\frac{2}{3}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.3.2

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

$f'' (x) = - \frac{{(8 - x^{3})}^{\frac{2}{3}} (2 x) - x^{2} \frac{d}{d x} {(8 - x^{3})}^{\frac{2}{3}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.3.3

Move $2$ to the left of ${(8 - x^{3})}^{\frac{2}{3}}$ .

$f'' (x) = - \frac{2 \cdot ({(8 - x^{3})}^{\frac{2}{3}} x) - x^{2} \frac{d}{d x} {(8 - x^{3})}^{\frac{2}{3}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{2}{3}}$ and $g (x) = 8 - x^{3}$ .

Tap for more steps...

Step 3.4.1

To apply the Chain Rule, set $u$ as $8 - x^{3}$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - x^{2} (\frac{d}{d u} (u^{\frac{2}{3}}) \frac{d}{d x} (8 - x^{3}))}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.4.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{2}{3}$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - x^{2} (\frac{2}{3} u^{\frac{2}{3} - 1} \frac{d}{d x} (8 - x^{3}))}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.4.3

Replace all occurrences of $u$ with $8 - x^{3}$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - x^{2} (\frac{2}{3} \cdot {(8 - x^{3})}^{\frac{2}{3} - 1} \frac{d}{d x} (8 - x^{3}))}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.5

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - x^{2} (\frac{2}{3} \cdot {(8 - x^{3})}^{\frac{2}{3} - 1 \cdot \frac{3}{3}} \frac{d}{d x} (8 - x^{3}))}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.6

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

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - x^{2} (\frac{2}{3} \cdot {(8 - x^{3})}^{\frac{2}{3} + \frac{- 1 \cdot 3}{3}} \frac{d}{d x} (8 - x^{3}))}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.7

Combine the numerators over the common denominator.

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - x^{2} (\frac{2}{3} \cdot {(8 - x^{3})}^{\frac{2 - 1 \cdot 3}{3}} \frac{d}{d x} (8 - x^{3}))}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.8

Simplify the numerator.

Tap for more steps...

Step 3.8.1

Multiply $- 1$ by $3$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - x^{2} (\frac{2}{3} \cdot {(8 - x^{3})}^{\frac{2 - 3}{3}} \frac{d}{d x} (8 - x^{3}))}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.8.2

Subtract $3$ from $2$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - x^{2} (\frac{2}{3} \cdot {(8 - x^{3})}^{\frac{- 1}{3}} \frac{d}{d x} (8 - x^{3}))}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.9

Combine fractions.

Tap for more steps...

Step 3.9.1

Move the negative in front of the fraction.

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - x^{2} (\frac{2}{3} \cdot {(8 - x^{3})}^{- \frac{1}{3}} \frac{d}{d x} (8 - x^{3}))}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.9.2

Combine $\frac{2}{3}$ and ${(8 - x^{3})}^{- \frac{1}{3}}$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - x^{2} (\frac{2 {(8 - x^{3})}^{- \frac{1}{3}}}{3} \cdot \frac{d}{d x} (8 - x^{3}))}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.9.3

Move ${(8 - x^{3})}^{- \frac{1}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - x^{2} (\frac{2}{3 {(8 - x^{3})}^{\frac{1}{3}}} \cdot \frac{d}{d x} (8 - x^{3}))}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.9.4

Combine $\frac{2}{3 {(8 - x^{3})}^{\frac{1}{3}}}$ and $x^{2}$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - \frac{2 x^{2}}{3 {(8 - x^{3})}^{\frac{1}{3}}} \cdot \frac{d}{d x} (8 - x^{3})}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.10

By the Sum Rule, the derivative of $8 - x^{3}$ with respect to $x$ is $\frac{d}{d x} [8] + \frac{d}{d x} [- x^{3}]$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - \frac{2 x^{2}}{3 {(8 - x^{3})}^{\frac{1}{3}}} \cdot (\frac{d}{d x} (8) + \frac{d}{d x} (- x^{3}))}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.11

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - \frac{2 x^{2}}{3 {(8 - x^{3})}^{\frac{1}{3}}} \cdot (0 + \frac{d}{d x} (- x^{3}))}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.12

Add $0$ and $\frac{d}{d x} [- x^{3}]$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - \frac{2 x^{2}}{3 {(8 - x^{3})}^{\frac{1}{3}}} \cdot \frac{d}{d x} (- x^{3})}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.13

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

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x - \frac{2 x^{2}}{3 {(8 - x^{3})}^{\frac{1}{3}}} \cdot (- \frac{d}{d x} x^{3})}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.14

Multiply.

Tap for more steps...

Step 3.14.1

Multiply $- 1$ by $- 1$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x + 1 (\frac{2 x^{2}}{3 {(8 - x^{3})}^{\frac{1}{3}}}) \cdot \frac{d}{d x} (x^{3})}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.14.2

Multiply $\frac{2 x^{2}}{3 {(8 - x^{3})}^{\frac{1}{3}}}$ by $1$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x + \frac{2 x^{2}}{3 {(8 - x^{3})}^{\frac{1}{3}}} \cdot \frac{d}{d x} (x^{3})}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.15

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

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x + \frac{2 x^{2}}{3 {(8 - x^{3})}^{\frac{1}{3}}} \cdot (3 x^{2})}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.16

Combine fractions.

Tap for more steps...

Step 3.16.1

Combine $3$ and $\frac{2 x^{2}}{3 {(8 - x^{3})}^{\frac{1}{3}}}$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x + \frac{3 (2 x^{2})}{3 {(8 - x^{3})}^{\frac{1}{3}}} \cdot x^{2}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.16.2

Multiply $2$ by $3$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x + \frac{6 x^{2}}{3 {(8 - x^{3})}^{\frac{1}{3}}} \cdot x^{2}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.16.3

Combine $\frac{6 x^{2}}{3 {(8 - x^{3})}^{\frac{1}{3}}}$ and $x^{2}$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x + \frac{6 x^{2} x^{2}}{3 {(8 - x^{3})}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.17

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

Tap for more steps...

Step 3.17.1

Move $x^{2}$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x + \frac{6 (x^{2} x^{2})}{3 {(8 - x^{3})}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.17.2

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

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x + \frac{6 x^{2 + 2}}{3 {(8 - x^{3})}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.17.3

Add $2$ and $2$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x + \frac{6 x^{4}}{3 {(8 - x^{3})}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.18

Factor $3$ out of $6 x^{4}$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x + \frac{3 (2 x^{4})}{3 {(8 - x^{3})}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.19

Cancel the common factors.

Tap for more steps...

Step 3.19.1

Factor $3$ out of $3 {(8 - x^{3})}^{\frac{1}{3}}$ .

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x + \frac{3 (2 x^{4})}{3 ({(8 - x^{3})}^{\frac{1}{3}})}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.19.2

Cancel the common factor.

Step 3.19.3

Rewrite the expression.

$f'' (x) = - \frac{2 {(8 - x^{3})}^{\frac{2}{3}} x + \frac{2 x^{4}}{{(8 - x^{3})}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.20

Reorder $8$ and $- x^{3}$ .

$f'' (x) = - \frac{2 x {(- x^{3} + 8)}^{\frac{2}{3}} + \frac{2 x^{4}}{{(- x^{3} + 8)}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.21

To write $2 x {(- x^{3} + 8)}^{\frac{2}{3}}$ as a fraction with a common denominator, multiply by $\frac{{(- x^{3} + 8)}^{\frac{1}{3}}}{{(- x^{3} + 8)}^{\frac{1}{3}}}$ .

$f'' (x) = - \frac{\frac{2 x {(- x^{3} + 8)}^{\frac{2}{3}} {(- x^{3} + 8)}^{\frac{1}{3}}}{{(- x^{3} + 8)}^{\frac{1}{3}}} + \frac{2 x^{4}}{{(- x^{3} + 8)}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.22

Combine the numerators over the common denominator.

$f'' (x) = - \frac{\frac{2 x {(- x^{3} + 8)}^{\frac{2}{3}} {(- x^{3} + 8)}^{\frac{1}{3}} + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.23

Multiply ${(- x^{3} + 8)}^{\frac{2}{3}}$ by ${(- x^{3} + 8)}^{\frac{1}{3}}$ by adding the exponents.

Tap for more steps...

Step 3.23.1

Move ${(- x^{3} + 8)}^{\frac{1}{3}}$ .

$f'' (x) = - \frac{\frac{2 x ({(- x^{3} + 8)}^{\frac{1}{3}} {(- x^{3} + 8)}^{\frac{2}{3}}) + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.23.2

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

$f'' (x) = - \frac{\frac{2 x {(- x^{3} + 8)}^{\frac{1}{3} + \frac{2}{3}} + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.23.3

Combine the numerators over the common denominator.

$f'' (x) = - \frac{\frac{2 x {(- x^{3} + 8)}^{\frac{1 + 2}{3}} + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.23.4

Add $1$ and $2$ .

$f'' (x) = - \frac{\frac{2 x {(- x^{3} + 8)}^{\frac{3}{3}} + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.23.5

Divide $3$ by $3$ .

$f'' (x) = - \frac{\frac{2 x (- x^{3} + 8) + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.24

Simplify $2 x {(- x^{3} + 8)}^{1}$ .

$f'' (x) = - \frac{\frac{2 x (- x^{3} + 8) + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.25

Rewrite $\frac{\frac{2 x (- x^{3} + 8) + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{1}{3}}}}{{(8 - x^{3})}^{\frac{4}{3}}}$ as a product.

$f'' (x) = - (\frac{2 x (- x^{3} + 8) + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{1}{3}}} \cdot \frac{1}{{(8 - x^{3})}^{\frac{4}{3}}}) + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.26

Multiply $\frac{2 x (- x^{3} + 8) + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{1}{3}}}$ by $\frac{1}{{(8 - x^{3})}^{\frac{4}{3}}}$ .

$f'' (x) = - \frac{2 x (- x^{3} + 8) + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{1}{3}} {(8 - x^{3})}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.27

Reorder terms.

$f'' (x) = - \frac{2 x (- x^{3} + 8) + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{1}{3}} {(- x^{3} + 8)}^{\frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.28

Multiply ${(- x^{3} + 8)}^{\frac{1}{3}}$ by ${(- x^{3} + 8)}^{\frac{4}{3}}$ by adding the exponents.

Tap for more steps...

Step 3.28.1

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

$f'' (x) = - \frac{2 x (- x^{3} + 8) + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{1}{3} + \frac{4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.28.2

Combine the numerators over the common denominator.

$f'' (x) = - \frac{2 x (- x^{3} + 8) + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{1 + 4}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.28.3

Add $1$ and $4$ .

$f'' (x) = - \frac{2 x (- x^{3} + 8) + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{5}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot \frac{d}{d x} (- 1)$

Step 3.29

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

$f'' (x) = - \frac{2 x (- x^{3} + 8) + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{5}{3}}} + \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} \cdot 0$

Step 3.30

Simplify the expression.

Tap for more steps...

Step 3.30.1

Multiply $\frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}}$ by $0$ .

$f'' (x) = - \frac{2 x (- x^{3} + 8) + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{5}{3}}} + 0$

Step 3.30.2

Add $- \frac{2 x (- x^{3} + 8) + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{5}{3}}}$ and $0$ .

$f'' (x) = - \frac{2 x (- x^{3} + 8) + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{5}{3}}}$

Step 3.31

Simplify.

Tap for more steps...

Step 3.31.1

Apply the distributive property.

$f'' (x) = - \frac{2 x (- x^{3}) + 2 x \cdot 8 + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{5}{3}}}$

Step 3.31.2

Simplify the numerator.

Tap for more steps...

Step 3.31.2.1

Simplify each term.

Tap for more steps...

Step 3.31.2.1.1

Rewrite using the commutative property of multiplication.

$f'' (x) = - \frac{2 \cdot (- 1 x \cdot x^{3}) + 2 x \cdot 8 + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{5}{3}}}$

Step 3.31.2.1.2

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

Tap for more steps...

Step 3.31.2.1.2.1

Move $x^{3}$ .

$f'' (x) = - \frac{2 \cdot (- 1 (x^{3} x)) + 2 x \cdot 8 + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{5}{3}}}$

Step 3.31.2.1.2.2

Multiply $x^{3}$ by $x$ .

Tap for more steps...

Step 3.31.2.1.2.2.1

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

$f'' (x) = - \frac{2 \cdot (- 1 (x^{3} x)) + 2 x \cdot 8 + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{5}{3}}}$

Step 3.31.2.1.2.2.2

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

$f'' (x) = - \frac{2 \cdot (- 1 x^{3 + 1}) + 2 x \cdot 8 + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{5}{3}}}$

Step 3.31.2.1.2.3

Add $3$ and $1$ .

$f'' (x) = - \frac{2 \cdot (- 1 x^{4}) + 2 x \cdot 8 + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{5}{3}}}$

Step 3.31.2.1.3

Multiply $2$ by $- 1$ .

$f'' (x) = - \frac{- 2 x^{4} + 2 x \cdot 8 + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{5}{3}}}$

Step 3.31.2.1.4

Multiply $8$ by $2$ .

$f'' (x) = - \frac{- 2 x^{4} + 16 x + 2 x^{4}}{{(- x^{3} + 8)}^{\frac{5}{3}}}$

Step 3.31.2.2

Combine the opposite terms in $- 2 x^{4} + 16 x + 2 x^{4}$ .

Tap for more steps...

Step 3.31.2.2.1

Add $- 2 x^{4}$ and $2 x^{4}$ .

$f'' (x) = - \frac{16 x + 0}{{(- x^{3} + 8)}^{\frac{5}{3}}}$

Step 3.31.2.2.2

Add $16 x$ and $0$ .

$f'' (x) = - \frac{16 x}{{(- x^{3} + 8)}^{\frac{5}{3}}}$

Step 4

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$- \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} = 0$

Step 5

Find the first derivative.

Tap for more steps...

Step 5.1

Find the first derivative.

Tap for more steps...

Step 5.1.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{8 - x^{3}}$ as ${(8 - x^{3})}^{\frac{1}{3}}$ .

$\frac{d}{d x} [{(8 - x^{3})}^{\frac{1}{3}}]$

Step 5.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{3}}$ and $g (x) = 8 - x^{3}$ .

Tap for more steps...

Step 5.1.2.1

To apply the Chain Rule, set $u$ as $8 - x^{3}$ .

$\frac{d}{d u} [u^{\frac{1}{3}}] \frac{d}{d x} [8 - x^{3}]$

Step 5.1.2.2

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

$\frac{1}{3} u^{\frac{1}{3} - 1} \frac{d}{d x} [8 - x^{3}]$

Step 5.1.2.3

Replace all occurrences of $u$ with $8 - x^{3}$ .

$\frac{1}{3} {(8 - x^{3})}^{\frac{1}{3} - 1} \frac{d}{d x} [8 - x^{3}]$

Step 5.1.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{1}{3} {(8 - x^{3})}^{\frac{1}{3} - 1 \cdot \frac{3}{3}} \frac{d}{d x} [8 - x^{3}]$

Step 5.1.4

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

$\frac{1}{3} {(8 - x^{3})}^{\frac{1}{3} + \frac{- 1 \cdot 3}{3}} \frac{d}{d x} [8 - x^{3}]$

Step 5.1.5

Combine the numerators over the common denominator.

$\frac{1}{3} {(8 - x^{3})}^{\frac{1 - 1 \cdot 3}{3}} \frac{d}{d x} [8 - x^{3}]$

Step 5.1.6

Simplify the numerator.

Tap for more steps...

Step 5.1.6.1

Multiply $- 1$ by $3$ .

$\frac{1}{3} {(8 - x^{3})}^{\frac{1 - 3}{3}} \frac{d}{d x} [8 - x^{3}]$

Step 5.1.6.2

Subtract $3$ from $1$ .

$\frac{1}{3} {(8 - x^{3})}^{\frac{- 2}{3}} \frac{d}{d x} [8 - x^{3}]$

Step 5.1.7

Combine fractions.

Tap for more steps...

Step 5.1.7.1

Move the negative in front of the fraction.

$\frac{1}{3} {(8 - x^{3})}^{- \frac{2}{3}} \frac{d}{d x} [8 - x^{3}]$

Step 5.1.7.2

Combine $\frac{1}{3}$ and ${(8 - x^{3})}^{- \frac{2}{3}}$ .

$\frac{{(8 - x^{3})}^{- \frac{2}{3}}}{3} \frac{d}{d x} [8 - x^{3}]$

Step 5.1.7.3

Move ${(8 - x^{3})}^{- \frac{2}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}} \frac{d}{d x} [8 - x^{3}]$

Step 5.1.8

By the Sum Rule, the derivative of $8 - x^{3}$ with respect to $x$ is $\frac{d}{d x} [8] + \frac{d}{d x} [- x^{3}]$ .

$\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}} (\frac{d}{d x} [8] + \frac{d}{d x} [- x^{3}])$

Step 5.1.9

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}} (0 + \frac{d}{d x} [- x^{3}])$

Step 5.1.10

Add $0$ and $\frac{d}{d x} [- x^{3}]$ .

$\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}} \frac{d}{d x} [- x^{3}]$

Step 5.1.11

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

$\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}} (- \frac{d}{d x} [x^{3}])$

Step 5.1.12

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

$\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}} (- (3 x^{2}))$

Step 5.1.13

Simplify terms.

Tap for more steps...

Step 5.1.13.1

Multiply $3$ by $- 1$ .

$\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}} (- 3 x^{2})$

Step 5.1.13.2

Combine $- 3$ and $\frac{1}{3 {(8 - x^{3})}^{\frac{2}{3}}}$ .

$\frac{- 3}{3 {(8 - x^{3})}^{\frac{2}{3}}} x^{2}$

Step 5.1.13.3

Combine $\frac{- 3}{3 {(8 - x^{3})}^{\frac{2}{3}}}$ and $x^{2}$ .

$\frac{- 3 x^{2}}{3 {(8 - x^{3})}^{\frac{2}{3}}}$

Step 5.1.13.4

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

$\frac{3 (- x^{2})}{3 {(8 - x^{3})}^{\frac{2}{3}}}$

Step 5.1.14

Cancel the common factors.

Tap for more steps...

Step 5.1.14.1

Factor $3$ out of $3 {(8 - x^{3})}^{\frac{2}{3}}$ .

$\frac{3 (- x^{2})}{3 ({(8 - x^{3})}^{\frac{2}{3}})}$

Step 5.1.14.2

Cancel the common factor.

$\frac{3 (- x^{2})}{3 {(8 - x^{3})}^{\frac{2}{3}}}$

Step 5.1.14.3

Rewrite the expression.

$\frac{- x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}}$

Step 5.1.15

Move the negative in front of the fraction.

$f' (x) = - \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}}$

Step 5.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}}$ .

$- \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}}$

Step 6

Set the first derivative equal to $0$ then solve the equation $- \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} = 0$ .

Tap for more steps...

Step 6.1

Set the first derivative equal to $0$ .

$- \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}} = 0$

Step 6.2

Set the numerator equal to zero.

$x^{2} = 0$

Step 6.3

Solve the equation for $x$ .

Tap for more steps...

Step 6.3.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Step 6.3.2

Simplify $\pm \sqrt{0}$ .

Tap for more steps...

Step 6.3.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 6.3.2.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Step 6.3.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 7

Find the values where the derivative is undefined.

Tap for more steps...

Step 7.1

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$- \frac{x^{2}}{\sqrt[3]{{(8 - x^{3})}^{2}}}$

Step 7.2

Set the denominator in $\frac{x^{2}}{\sqrt[3]{{(8 - x^{3})}^{2}}}$ equal to $0$ to find where the expression is undefined.

$\sqrt[3]{{(8 - x^{3})}^{2}} = 0$

Step 7.3

Solve for $x$ .

Tap for more steps...

Step 7.3.1

To remove the radical on the left side of the equation, cube both sides of the equation.

${\sqrt[3]{{(8 - x^{3})}^{2}}}^{3} = 0^{3}$

Step 7.3.2

Simplify each side of the equation.

Tap for more steps...

Step 7.3.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{{(8 - x^{3})}^{2}}$ as ${(8 - x^{3})}^{\frac{2}{3}}$ .

${({(8 - x^{3})}^{\frac{2}{3}})}^{3} = 0^{3}$

Step 7.3.2.2

Simplify the left side.

Tap for more steps...

Step 7.3.2.2.1

Multiply the exponents in ${({(8 - x^{3})}^{\frac{2}{3}})}^{3}$ .

Tap for more steps...

Step 7.3.2.2.1.1

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

${(8 - x^{3})}^{\frac{2}{3} \cdot 3} = 0^{3}$

Step 7.3.2.2.1.2

Cancel the common factor of $3$ .

Tap for more steps...

Step 7.3.2.2.1.2.1

Cancel the common factor.

${(8 - x^{3})}^{\frac{2}{3} \cdot 3} = 0^{3}$

Step 7.3.2.2.1.2.2

Rewrite the expression.

${(8 - x^{3})}^{2} = 0^{3}$

Step 7.3.2.3

Simplify the right side.

Tap for more steps...

Step 7.3.2.3.1

Raising $0$ to any positive power yields $0$ .

${(8 - x^{3})}^{2} = 0$

Step 7.3.3

Solve for $x$ .

Tap for more steps...

Step 7.3.3.1

Factor the left side of the equation.

Tap for more steps...

Step 7.3.3.1.1

Rewrite $8$ as $2^{3}$ .

${(2^{3} - x^{3})}^{2} = 0$

Step 7.3.3.1.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = 2$ and $b = x$ .

${((2 - x) (2^{2} + 2 x + x^{2}))}^{2} = 0$

Step 7.3.3.1.3

Raise $2$ to the power of $2$ .

${((2 - x) (4 + 2 x + x^{2}))}^{2} = 0$

Step 7.3.3.1.4

Apply the product rule to $(2 - x) (4 + 2 x + x^{2})$ .

${(2 - x)}^{2} {(4 + 2 x + x^{2})}^{2} = 0$

Step 7.3.3.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

${(2 - x)}^{2} = 0$

${(4 + 2 x + x^{2})}^{2} = 0$

Step 7.3.3.3

Set ${(2 - x)}^{2}$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 7.3.3.3.1

Set ${(2 - x)}^{2}$ equal to $0$ .

${(2 - x)}^{2} = 0$

Step 7.3.3.3.2

Solve ${(2 - x)}^{2} = 0$ for $x$ .

Tap for more steps...

Step 7.3.3.3.2.1

Set the $2 - x$ equal to $0$ .

$2 - x = 0$

Step 7.3.3.3.2.2

Solve for $x$ .

Tap for more steps...

Step 7.3.3.3.2.2.1

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 7.3.3.3.2.2.2

Divide each term in $- x = - 2$ by $- 1$ and simplify.

Tap for more steps...

Step 7.3.3.3.2.2.2.1

Divide each term in $- x = - 2$ by $- 1$ .

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

Step 7.3.3.3.2.2.2.2

Simplify the left side.

Tap for more steps...

Step 7.3.3.3.2.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 7.3.3.3.2.2.2.2.2

Divide $x$ by $1$ .

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

Step 7.3.3.3.2.2.2.3

Simplify the right side.

Tap for more steps...

Step 7.3.3.3.2.2.2.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 7.3.3.4

Set ${(4 + 2 x + x^{2})}^{2}$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 7.3.3.4.1

Set ${(4 + 2 x + x^{2})}^{2}$ equal to $0$ .

${(4 + 2 x + x^{2})}^{2} = 0$

Step 7.3.3.4.2

Solve ${(4 + 2 x + x^{2})}^{2} = 0$ for $x$ .

Tap for more steps...

Step 7.3.3.4.2.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$4 + 2 x + x^{2} = \pm \sqrt{0}$

Step 7.3.3.4.2.2

Simplify $\pm \sqrt{0}$ .

Tap for more steps...

Step 7.3.3.4.2.2.1

Rewrite $0$ as $0^{2}$ .

$4 + 2 x + x^{2} = \pm \sqrt{0^{2}}$

Step 7.3.3.4.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$4 + 2 x + x^{2} = \pm 0$

Step 7.3.3.4.2.2.3

Plus or minus $0$ is $0$ .

$4 + 2 x + x^{2} = 0$

Step 7.3.3.4.2.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 7.3.3.4.2.4

Substitute the values $a = 1$ , $b = 2$ , and $c = 4$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot 4)}}{2 \cdot 1}$

Step 7.3.3.4.2.5

Simplify.

Tap for more steps...

Step 7.3.3.4.2.5.1

Simplify the numerator.

Tap for more steps...

Step 7.3.3.4.2.5.1.1

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Step 7.3.3.4.2.5.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

Tap for more steps...

Step 7.3.3.4.2.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

Step 7.3.3.4.2.5.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Step 7.3.3.4.2.5.1.3

Subtract $16$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 7.3.3.4.2.5.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 7.3.3.4.2.5.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 7.3.3.4.2.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 7.3.3.4.2.5.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

Tap for more steps...

Step 7.3.3.4.2.5.1.7.1

Factor $4$ out of $12$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Step 7.3.3.4.2.5.1.7.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 7.3.3.4.2.5.1.8

Pull terms out from under the radical.

$x = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

Step 7.3.3.4.2.5.1.9

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Step 7.3.3.4.2.5.2

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2}$

Step 7.3.3.4.2.5.3

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

$x = - 1 \pm i \sqrt{3}$

Step 7.3.3.4.2.6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Tap for more steps...

Step 7.3.3.4.2.6.1

Simplify the numerator.

Tap for more steps...

Step 7.3.3.4.2.6.1.1

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Step 7.3.3.4.2.6.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

Tap for more steps...

Step 7.3.3.4.2.6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

Step 7.3.3.4.2.6.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Step 7.3.3.4.2.6.1.3

Subtract $16$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 7.3.3.4.2.6.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 7.3.3.4.2.6.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 7.3.3.4.2.6.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 7.3.3.4.2.6.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

Tap for more steps...

Step 7.3.3.4.2.6.1.7.1

Factor $4$ out of $12$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Step 7.3.3.4.2.6.1.7.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 7.3.3.4.2.6.1.8

Pull terms out from under the radical.

$x = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

Step 7.3.3.4.2.6.1.9

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Step 7.3.3.4.2.6.2

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2}$

Step 7.3.3.4.2.6.3

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

$x = - 1 \pm i \sqrt{3}$

Step 7.3.3.4.2.6.4

Change the $\pm$ to $+$ .

$x = - 1 + i \sqrt{3}$

Step 7.3.3.4.2.7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Tap for more steps...

Step 7.3.3.4.2.7.1

Simplify the numerator.

Tap for more steps...

Step 7.3.3.4.2.7.1.1

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Step 7.3.3.4.2.7.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

Tap for more steps...

Step 7.3.3.4.2.7.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

Step 7.3.3.4.2.7.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Step 7.3.3.4.2.7.1.3

Subtract $16$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 7.3.3.4.2.7.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 7.3.3.4.2.7.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 7.3.3.4.2.7.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 7.3.3.4.2.7.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

Tap for more steps...

Step 7.3.3.4.2.7.1.7.1

Factor $4$ out of $12$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Step 7.3.3.4.2.7.1.7.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 7.3.3.4.2.7.1.8

Pull terms out from under the radical.

$x = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

Step 7.3.3.4.2.7.1.9

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Step 7.3.3.4.2.7.2

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2}$

Step 7.3.3.4.2.7.3

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

$x = - 1 \pm i \sqrt{3}$

Step 7.3.3.4.2.7.4

Change the $\pm$ to $-$ .

$x = - 1 - i \sqrt{3}$

Step 7.3.3.4.2.8

The final answer is the combination of both solutions.

$x = - 1 + i \sqrt{3}, - 1 - i \sqrt{3}$

Step 7.3.3.5

The final solution is all the values that make ${(2 - x)}^{2} {(4 + 2 x + x^{2})}^{2} = 0$ true.

$x = 2, - 1 + i \sqrt{3}, - 1 - i \sqrt{3}$

Step 7.4

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = 2$

Step 8

Critical points to evaluate.

$x = 0, 2$

Step 9

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{16 (0)}{{(- {(0)}^{3} + 8)}^{\frac{5}{3}}}$

Step 10

Evaluate the second derivative.

Tap for more steps...

Step 10.1

Multiply $16$ by $0$ .

$- \frac{0}{{(- 0^{3} + 8)}^{\frac{5}{3}}}$

Step 10.2

Simplify the denominator.

Tap for more steps...

Step 10.2.1

Simplify each term.

Tap for more steps...

Step 10.2.1.1

Raising $0$ to any positive power yields $0$ .

$- \frac{0}{{(- 0 + 8)}^{\frac{5}{3}}}$

Step 10.2.1.2

Multiply $- 1$ by $0$ .

$- \frac{0}{{(0 + 8)}^{\frac{5}{3}}}$

Step 10.2.2

Add $0$ and $8$ .

$- \frac{0}{8^{\frac{5}{3}}}$

Step 10.2.3

Rewrite $8$ as $2^{3}$ .

$- \frac{0}{{(2^{3})}^{\frac{5}{3}}}$

Step 10.2.4

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

$- \frac{0}{2^{3 (\frac{5}{3})}}$

Step 10.2.5

Cancel the common factor of $3$ .

Tap for more steps...

Step 10.2.5.1

Cancel the common factor.

$- \frac{0}{2^{3 (\frac{5}{3})}}$

Step 10.2.5.2

Rewrite the expression.

$- \frac{0}{2^{5}}$

Step 10.2.6

Raise $2$ to the power of $5$ .

$- \frac{0}{32}$

Step 10.3

Simplify the expression.

Tap for more steps...

Step 10.3.1

Divide $0$ by $32$ .

$- 0$

Step 10.3.2

Multiply $- 1$ by $0$ .

$0$

Step 11

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

Tap for more steps...

Step 11.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, 0) \cup (0, 2) \cup (2, \infty)$

Step 11.2

Substitute any number, such as $- 2$ , from the interval $(- \infty, 0)$ in the first derivative $- \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}}$ to check if the result is negative or positive.

Tap for more steps...

Step 11.2.1

Replace the variable $x$ with $- 2$ in the expression.

$f' (- 2) = - \frac{{(- 2)}^{2}}{{(8 - {(- 2)}^{3})}^{\frac{2}{3}}}$

Step 11.2.2

Simplify the result.

Tap for more steps...

Step 11.2.2.1

Raise $- 2$ to the power of $2$ .

$f' (- 2) = - \frac{4}{{(8 - {(- 2)}^{3})}^{\frac{2}{3}}}$

Step 11.2.2.2

Simplify the denominator.

Tap for more steps...

Step 11.2.2.2.1

Simplify each term.

Tap for more steps...

Step 11.2.2.2.1.1

Raise $- 2$ to the power of $3$ .

$f' (- 2) = - \frac{4}{{(8 + 8)}^{\frac{2}{3}}}$

Step 11.2.2.2.1.2

Multiply $- 1$ by $- 8$ .

$f' (- 2) = - \frac{4}{{(8 + 8)}^{\frac{2}{3}}}$

Step 11.2.2.2.2

Add $8$ and $8$ .

$f' (- 2) = - \frac{4}{16^{\frac{2}{3}}}$

Step 11.2.2.3

The final answer is $- \frac{4}{16^{\frac{2}{3}}}$ .

$- \frac{4}{16^{\frac{2}{3}}}$

Step 11.3

Substitute any number, such as $1$ , from the interval $(0, 2)$ in the first derivative $- \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}}$ to check if the result is negative or positive.

Tap for more steps...

Step 11.3.1

Replace the variable $x$ with $1$ in the expression.

$f' (1) = - \frac{{(1)}^{2}}{{(8 - {(1)}^{3})}^{\frac{2}{3}}}$

Step 11.3.2

Simplify the result.

Tap for more steps...

Step 11.3.2.1

One to any power is one.

$f' (1) = - \frac{1}{{(8 - 1^{3})}^{\frac{2}{3}}}$

Step 11.3.2.2

Simplify the denominator.

Tap for more steps...

Step 11.3.2.2.1

Simplify each term.

Tap for more steps...

Step 11.3.2.2.1.1

One to any power is one.

$f' (1) = - \frac{1}{{(8 - 1 \cdot 1)}^{\frac{2}{3}}}$

Step 11.3.2.2.1.2

Multiply $- 1$ by $1$ .

$f' (1) = - \frac{1}{{(8 - 1)}^{\frac{2}{3}}}$

Step 11.3.2.2.2

Subtract $1$ from $8$ .

$f' (1) = - \frac{1}{7^{\frac{2}{3}}}$

Step 11.3.2.3

The final answer is $- \frac{1}{7^{\frac{2}{3}}}$ .

$- \frac{1}{7^{\frac{2}{3}}}$

Step 11.4

Substitute any number, such as $4$ , from the interval $(2, \infty)$ in the first derivative $- \frac{x^{2}}{{(8 - x^{3})}^{\frac{2}{3}}}$ to check if the result is negative or positive.

Tap for more steps...

Step 11.4.1

Replace the variable $x$ with $4$ in the expression.

$f' (4) = - \frac{{(4)}^{2}}{{(8 - {(4)}^{3})}^{\frac{2}{3}}}$

Step 11.4.2

Simplify the result.

Tap for more steps...

Step 11.4.2.1

Raise $4$ to the power of $2$ .

$f' (4) = - \frac{16}{{(8 - 4^{3})}^{\frac{2}{3}}}$

Step 11.4.2.2

Simplify the denominator.

Tap for more steps...

Step 11.4.2.2.1

Simplify each term.

Tap for more steps...

Step 11.4.2.2.1.1

Raise $4$ to the power of $3$ .

$f' (4) = - \frac{16}{{(8 - 1 \cdot 64)}^{\frac{2}{3}}}$

Step 11.4.2.2.1.2

Multiply $- 1$ by $64$ .

$f' (4) = - \frac{16}{{(8 - 64)}^{\frac{2}{3}}}$

Step 11.4.2.2.2

Subtract $64$ from $8$ .

$f' (4) = - \frac{16}{{(- 56)}^{\frac{2}{3}}}$

Step 11.4.2.3

The final answer is $- \frac{16}{{(- 56)}^{\frac{2}{3}}}$ .

$- \frac{16}{{(- 56)}^{\frac{2}{3}}}$

Step 11.5

Since the first derivative did not change signs around $x = 0$ , this is not a local maximum or minimum.

Not a local maximum or minimum

Step 11.6

No local maxima or minima found for $f (x) = \sqrt[3]{8 - x^{3}}$ .

No local maxima or minima

Step 12