Calculus Examples

$f (x) = {(\sqrt[4]{x^{2} + 1})}^{3}$

Step 1

Rewrite ${(\sqrt[4]{x^{2} + 1})}^{3}$ as $\sqrt[4]{{(x^{2} + 1)}^{3}}$ .

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

Step 2

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[4]{{(x^{2} + 1)}^{3}}$ as ${(x^{2} + 1)}^{\frac{3}{4}}$ .

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

Step 3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{3}{4}}$ and $g (x) = x^{2} + 1$ .

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Step 3.1

To apply the Chain Rule, set $u$ as $x^{2} + 1$ .

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

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{3}{4}$ .

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

Step 3.3

Replace all occurrences of $u$ with $x^{2} + 1$ .

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

Step 4

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Step 5

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

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

Step 6

Combine the numerators over the common denominator.

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

Step 7

Simplify the numerator.

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Step 7.1

Multiply $- 1$ by $4$ .

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

Step 7.2

Subtract $4$ from $3$ .

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

Step 8

Combine fractions.

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Step 8.1

Move the negative in front of the fraction.

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

Step 8.2

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

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

Step 8.3

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

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

Step 9

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

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

Step 10

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

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

Step 11

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

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

Step 12

Simplify terms.

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Step 12.1

Add $2 x$ and $0$ .

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

Step 12.2

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

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

Step 12.3

Multiply $2$ by $3$ .

$\frac{6}{4 {(x^{2} + 1)}^{\frac{1}{4}}} x$

Step 12.4

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

$\frac{6 x}{4 {(x^{2} + 1)}^{\frac{1}{4}}}$

Step 12.5

Factor $2$ out of $6 x$ .

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

Step 13

Cancel the common factors.

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Step 13.1

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

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

Step 13.2

Cancel the common factor.

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

Step 13.3

Rewrite the expression.

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