Calculus Examples

$f' (x) = \sqrt{x^{4} + 1} + x^{3} - 3 x$

Find the first derivative of the function.

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By the Sum Rule, the derivative of $\sqrt{x^{4} + 1} + x^{3} - 3 x$ with respect to $x$ is $\frac{d}{d x} [\sqrt{x^{4} + 1}] + \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 3 x]$ .

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

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

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{4} + 1}$ as ${(x^{4} + 1)}^{\frac{1}{2}}$ .

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

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x^{4} + 1$ .

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To apply the Chain Rule, set $u$ as $x^{4} + 1$ .

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

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

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

Replace all occurrences of $u$ with $x^{4} + 1$ .

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

By the Sum Rule, the derivative of $x^{4} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [1]$ .

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

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

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

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

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

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

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Multiply $- 1$ by $2$ .

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

Subtract $2$ from $1$ .

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

Move the negative in front of the fraction.

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

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

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

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

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

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

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

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

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

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

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

Factor $2$ out of $4 x^{3}$ .

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

Cancel the common factors.

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Factor $2$ out of $2 {(x^{4} + 1)}^{\frac{1}{2}}$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

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

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

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

Multiply $- 3$ by $1$ .

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

Find the second derivative of the function.

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By the Sum Rule, the derivative of $\frac{2 x^{3}}{{(x^{4} + 1)}^{\frac{1}{2}}} + 3 x^{2} - 3$ with respect to $x$ is $\frac{d}{d x} [\frac{2 x^{3}}{{(x^{4} + 1)}^{\frac{1}{2}}}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 3]$ .

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

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

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Since $2$ is constant with respect to $x$ , the derivative of $\frac{2 x^{3}}{{(x^{4} + 1)}^{\frac{1}{2}}}$ with respect to $x$ is $2 \frac{d}{d x} [\frac{x^{3}}{{(x^{4} + 1)}^{\frac{1}{2}}}]$ .

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

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

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

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

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

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x^{4} + 1$ .

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To apply the Chain Rule, set $u$ as $x^{4} + 1$ .

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

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

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

Replace all occurrences of $u$ with $x^{4} + 1$ .

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

By the Sum Rule, the derivative of $x^{4} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [1]$ .

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

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

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

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

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

Move $3$ to the left of ${(x^{4} + 1)}^{\frac{1}{2}}$ .

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

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

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Multiply $- 1$ by $2$ .

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

Subtract $2$ from $1$ .

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

Move the negative in front of the fraction.

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

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

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

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

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

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

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

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

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

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

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

Factor $2$ out of $4 x^{3}$ .

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

Cancel the common factors.

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Factor $2$ out of $2 {(x^{4} + 1)}^{\frac{1}{2}}$ .

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

Cancel the common factor.

Rewrite the expression.

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

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

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

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

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Move $x^{3}$ .

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

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

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

Add $3$ and $3$ .

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

Combine $3 {(x^{4} + 1)}^{\frac{1}{2}} x^{2}$ and $- \frac{2 x^{6}}{{(x^{4} + 1)}^{\frac{1}{2}}}$ using a common denominator.

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Move ${(x^{4} + 1)}^{\frac{1}{2}}$ .

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

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

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

Combine the numerators over the common denominator.

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

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

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Move ${(x^{4} + 1)}^{\frac{1}{2}}$ .

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

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

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

Combine the numerators over the common denominator.

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

Add $1$ and $1$ .

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

Divide $2$ by $2$ .

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

Simplify $3 x^{2} {(x^{4} + 1)}^{1}$ .

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

Multiply the exponents in ${({(x^{4} + 1)}^{\frac{1}{2}})}^{2}$ .

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Rewrite $\frac{\frac{3 x^{2} (x^{4} + 1) - 2 x^{6}}{{(x^{4} + 1)}^{\frac{1}{2}}}}{x^{4} + 1}$ as a product.

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

Multiply $\frac{3 x^{2} (x^{4} + 1) - 2 x^{6}}{{(x^{4} + 1)}^{\frac{1}{2}}}$ by $\frac{1}{x^{4} + 1}$ .

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

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

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

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Raise $x^{4} + 1$ to the power of $1$ .

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

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

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

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Add $1$ and $2$ .

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

Combine $2$ and $\frac{3 x^{2} (x^{4} + 1) - 2 x^{6}}{{(x^{4} + 1)}^{\frac{3}{2}}}$ .

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

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

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

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

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

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

Multiply $2$ by $3$ .

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

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$f''' (x) = \frac{2 (3 x^{2} (x^{4} + 1) - 2 x^{6})}{{(x^{4} + 1)}^{\frac{3}{2}}} + 6 x + 0$

Simplify.

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Apply the distributive property.

$f''' (x) = \frac{2 (3 x^{2} x^{4} + 3 x^{2} \cdot 1 - 2 x^{6})}{{(x^{4} + 1)}^{\frac{3}{2}}} + 6 x + 0$

Apply the distributive property.

$f''' (x) = \frac{2 (3 x^{2} x^{4}) + 2 (3 x^{2} \cdot 1) + 2 (- 2 x^{6})}{{(x^{4} + 1)}^{\frac{3}{2}}} + 6 x + 0$

Combine terms.

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Multiply $x^{2}$ by $x^{4}$ by adding the exponents.

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Move $x^{4}$ .

$f''' (x) = \frac{2 (3 (x^{4} x^{2})) + 2 (3 x^{2} \cdot 1) + 2 (- 2 x^{6})}{{(x^{4} + 1)}^{\frac{3}{2}}} + 6 x + 0$

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

$f''' (x) = \frac{2 (3 x^{4 + 2}) + 2 (3 x^{2} \cdot 1) + 2 (- 2 x^{6})}{{(x^{4} + 1)}^{\frac{3}{2}}} + 6 x + 0$

Add $4$ and $2$ .

$f''' (x) = \frac{2 (3 x^{6}) + 2 (3 x^{2} \cdot 1) + 2 (- 2 x^{6})}{{(x^{4} + 1)}^{\frac{3}{2}}} + 6 x + 0$

Multiply $3$ by $2$ .

$f''' (x) = \frac{6 x^{6} + 2 (3 x^{2} \cdot 1) + 2 (- 2 x^{6})}{{(x^{4} + 1)}^{\frac{3}{2}}} + 6 x + 0$

Multiply $3$ by $1$ .

$f''' (x) = \frac{6 x^{6} + 2 (3 x^{2}) + 2 (- 2 x^{6})}{{(x^{4} + 1)}^{\frac{3}{2}}} + 6 x + 0$

Multiply $3$ by $2$ .

$f''' (x) = \frac{6 x^{6} + 6 x^{2} + 2 (- 2 x^{6})}{{(x^{4} + 1)}^{\frac{3}{2}}} + 6 x + 0$

Multiply $- 2$ by $2$ .

$f''' (x) = \frac{6 x^{6} + 6 x^{2} - 4 x^{6}}{{(x^{4} + 1)}^{\frac{3}{2}}} + 6 x + 0$

Subtract $4 x^{6}$ from $6 x^{6}$ .

$f''' (x) = \frac{2 x^{6} + 6 x^{2}}{{(x^{4} + 1)}^{\frac{3}{2}}} + 6 x + 0$

To write $6 x$ as a fraction with a common denominator, multiply by $\frac{{(x^{4} + 1)}^{\frac{3}{2}}}{{(x^{4} + 1)}^{\frac{3}{2}}}$ .

$f''' (x) = \frac{2 x^{6} + 6 x^{2}}{{(x^{4} + 1)}^{\frac{3}{2}}} + \frac{6 x {(x^{4} + 1)}^{\frac{3}{2}}}{{(x^{4} + 1)}^{\frac{3}{2}}} + 0$

Combine the numerators over the common denominator.

$f''' (x) = \frac{2 x^{6} + 6 x^{2} + 6 x {(x^{4} + 1)}^{\frac{3}{2}}}{{(x^{4} + 1)}^{\frac{3}{2}}} + 0$

Add $\frac{2 x^{6} + 6 x^{2} + 6 x {(x^{4} + 1)}^{\frac{3}{2}}}{{(x^{4} + 1)}^{\frac{3}{2}}}$ and $0$ .

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

Factor $2 x$ out of $2 x^{6} + 6 x^{2} + 6 x {(x^{4} + 1)}^{\frac{3}{2}}$ .

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Factor $2 x$ out of $2 x^{6}$ .

$f''' (x) = \frac{2 x (x^{5}) + 6 x^{2} + 6 x {(x^{4} + 1)}^{\frac{3}{2}}}{{(x^{4} + 1)}^{\frac{3}{2}}}$

Factor $2 x$ out of $6 x^{2}$ .

$f''' (x) = \frac{2 x (x^{5}) + 2 x (3 x) + 6 x {(x^{4} + 1)}^{\frac{3}{2}}}{{(x^{4} + 1)}^{\frac{3}{2}}}$

Factor $2 x$ out of $6 x {(x^{4} + 1)}^{\frac{3}{2}}$ .

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

Factor $2 x$ out of $2 x (x^{5}) + 2 x (3 x)$ .

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

Factor $2 x$ out of $2 x (x^{5} + 3 x) + 2 x (3 {(x^{4} + 1)}^{\frac{3}{2}})$ .

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

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

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

Find the first derivative.

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Find the first derivative.

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By the Sum Rule, the derivative of $\sqrt{x^{4} + 1} + x^{3} - 3 x$ with respect to $x$ is $\frac{d}{d x} [\sqrt{x^{4} + 1}] + \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 3 x]$ .

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

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

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{4} + 1}$ as ${(x^{4} + 1)}^{\frac{1}{2}}$ .

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

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x^{4} + 1$ .

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To apply the Chain Rule, set $u$ as $x^{4} + 1$ .

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

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

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

Replace all occurrences of $u$ with $x^{4} + 1$ .

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

By the Sum Rule, the derivative of $x^{4} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [1]$ .

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

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

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

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

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

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

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Multiply $- 1$ by $2$ .

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

Subtract $2$ from $1$ .

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

Move the negative in front of the fraction.

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

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

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

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

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

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

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

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

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

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

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

Factor $2$ out of $4 x^{3}$ .

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

Cancel the common factors.

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Factor $2$ out of $2 {(x^{4} + 1)}^{\frac{1}{2}}$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

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

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

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

Multiply $- 3$ by $1$ .

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

The first derivative of $f' (x)$ with respect to $x$ is $\frac{2 x^{3}}{{(x^{4} + 1)}^{\frac{1}{2}}} + 3 x^{2} - 3$ .

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

Set the first derivative equal to $0$ then solve the equation $\frac{2 x^{3}}{{(x^{4} + 1)}^{\frac{1}{2}}} + 3 x^{2} - 3 = 0$ .

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Set the first derivative equal to $0$ .

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

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x \approx - 1.3316463, 0.82889699$

Find the values where the derivative is undefined.

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The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Critical points to evaluate.

$x = - 1.3316463, 0.82889699$

Evaluate the second derivative at $x = - 1.3316463$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{2 (- 1.3316463) ({(- 1.3316463)}^{5} + 3 (- 1.3316463) + 3 {({(- 1.3316463)}^{4} + 1)}^{\frac{3}{2}})}{{({(- 1.3316463)}^{4} + 1)}^{\frac{3}{2}}}$

Evaluate the second derivative.

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Simplify the numerator.

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Multiply $2$ by $- 1.3316463$ .

$\frac{- 2.6632926 ({(- 1.3316463)}^{5} + 3 (- 1.3316463) + 3 {({(- 1.3316463)}^{4} + 1)}^{\frac{3}{2}})}{{({(- 1.3316463)}^{4} + 1)}^{\frac{3}{2}}}$

Multiply $- 2.6632926$ by ${(- 1.3316463)}^{5} + 3 (- 1.3316463) + 3 {({(- 1.3316463)}^{4} + 1)}^{\frac{3}{2}}$ .

$\frac{- 45.62246541}{{({(- 1.3316463)}^{4} + 1)}^{\frac{3}{2}}}$

Simplify the denominator.

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Raise $- 1.3316463$ to the power of $4$ .

$\frac{- 45.62246541}{{(3.14452862 + 1)}^{\frac{3}{2}}}$

Add $3.14452862$ and $1$ .

$\frac{- 45.62246541}{{4.14452862}^{\frac{3}{2}}}$

Rewrite $4.14452862$ as ${2.03581153}^{2}$ .

$\frac{- 45.62246541}{{({2.03581153}^{2})}^{\frac{3}{2}}}$

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

$\frac{- 45.62246541}{{2.03581153}^{2 (\frac{3}{2})}}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{- 45.62246541}{{2.03581153}^{2 (\frac{3}{2})}}$

Rewrite the expression.

$\frac{- 45.62246541}{{2.03581153}^{3}}$

Raise $2.03581153$ to the power of $3$ .

$\frac{- 45.62246541}{8.43747919}$

Divide $- 45.62246541$ by $8.43747919$ .

$- 5.40712034$

$x = - 1.3316463$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 1.3316463$ is a local maximum

Find the y-value when $x = - 1.3316463$ .

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Replace the variable $x$ with $- 1.3316463$ in the expression.

$f' (- 1.3316463) = \sqrt{{(- 1.3316463)}^{4} + 1} + {(- 1.3316463)}^{3} - 3 \cdot - 1.3316463$

Simplify the result.

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Simplify each term.

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Raise $- 1.3316463$ to the power of $4$ .

$f' (- 1.3316463) = \sqrt{3.14452862 + 1} + {(- 1.3316463)}^{3} - 3 \cdot - 1.3316463$

Add $3.14452862$ and $1$ .

$f' (- 1.3316463) = \sqrt{4.14452862} + {(- 1.3316463)}^{3} - 3 \cdot - 1.3316463$

Raise $- 1.3316463$ to the power of $3$ .

$f' (- 1.3316463) = \sqrt{4.14452862} - 2.36138426 - 3 \cdot - 1.3316463$

Multiply $- 3$ by $- 1.3316463$ .

$f' (- 1.3316463) = \sqrt{4.14452862} - 2.36138426 + 3.99493891$

Simplify by adding terms.

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Subtract $2.36138426$ from $\sqrt{4.14452862}$ .

$f' (- 1.3316463) = - 0.32557272 + 3.99493891$

Add $- 0.32557272$ and $3.99493891$ .

$f' (- 1.3316463) = 3.66936619$

The final answer is $3.66936619$ .

$y = 3.66936619$

Evaluate the second derivative at $x = 0.82889699$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{2 (0.82889699) ({(0.82889699)}^{5} + 3 (0.82889699) + 3 {({(0.82889699)}^{4} + 1)}^{\frac{3}{2}})}{{({(0.82889699)}^{4} + 1)}^{\frac{3}{2}}}$

Evaluate the second derivative.

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Simplify the numerator.

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Multiply $2$ by $0.82889699$ .

$\frac{1.65779399 ({0.82889699}^{5} + 3 (0.82889699) + 3 {({0.82889699}^{4} + 1)}^{\frac{3}{2}})}{{({0.82889699}^{4} + 1)}^{\frac{3}{2}}}$

Multiply $1.65779399$ by ${0.82889699}^{5} + 3 (0.82889699) + 3 {({0.82889699}^{4} + 1)}^{\frac{3}{2}}$ .

$\frac{13.65375517}{{({0.82889699}^{4} + 1)}^{\frac{3}{2}}}$

Simplify the denominator.

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Raise $0.82889699$ to the power of $4$ .

$\frac{13.65375517}{{(0.4720655 + 1)}^{\frac{3}{2}}}$

Add $0.4720655$ and $1$ .

$\frac{13.65375517}{{1.4720655}^{\frac{3}{2}}}$

Rewrite $1.4720655$ as ${1.21328706}^{2}$ .

$\frac{13.65375517}{{({1.21328706}^{2})}^{\frac{3}{2}}}$

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

$\frac{13.65375517}{{1.21328706}^{2 (\frac{3}{2})}}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{13.65375517}{{1.21328706}^{2 (\frac{3}{2})}}$

Rewrite the expression.

$\frac{13.65375517}{{1.21328706}^{3}}$

Raise $1.21328706$ to the power of $3$ .

$\frac{13.65375517}{1.78603804}$

Divide $13.65375517$ by $1.78603804$ .

$7.64471687$

$x = 0.82889699$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 0.82889699$ is a local minimum

Find the y-value when $x = 0.82889699$ .

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Replace the variable $x$ with $0.82889699$ in the expression.

$f' (0.82889699) = \sqrt{{(0.82889699)}^{4} + 1} + {(0.82889699)}^{3} - 3 \cdot 0.82889699$

Simplify the result.

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Simplify each term.

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Raise $0.82889699$ to the power of $4$ .

$f' (0.82889699) = \sqrt{0.4720655 + 1} + {(0.82889699)}^{3} - 3 \cdot 0.82889699$

Add $0.4720655$ and $1$ .

$f' (0.82889699) = \sqrt{1.4720655} + {(0.82889699)}^{3} - 3 \cdot 0.82889699$

Raise $0.82889699$ to the power of $3$ .

$f' (0.82889699) = \sqrt{1.4720655} + 0.56951045 - 3 \cdot 0.82889699$

Multiply $- 3$ by $0.82889699$ .

$f' (0.82889699) = \sqrt{1.4720655} + 0.56951045 - 2.48669099$

Simplify by adding terms.

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Add $\sqrt{1.4720655}$ and $0.56951045$ .

$f' (0.82889699) = 1.78279752 - 2.48669099$

Subtract $2.48669099$ from $1.78279752$ .

$f' (0.82889699) = - 0.70389347$

The final answer is $- 0.70389347$ .

$y = - 0.70389347$

These are the local extrema for $f' (x) = \sqrt{x^{4} + 1} + x^{3} - 3 x$ .

$(- 1.3316463, 3.66936619)$ is a local maxima

$(0.82889699, - 0.70389347)$ is a local minima