Calculus Examples

$y = {(2 \tan^{3} (2 x) - 1)}^{\frac{1}{3}}$

Step 1

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 $g (x) = 2 \tan^{3} (2 x) - 1$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $2 \tan^{3} (2 x) - 1$ .

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

Step 1.2

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

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

Step 1.3

Replace all occurrences of $u_{1}$ with $2 \tan^{3} (2 x) - 1$ .

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

Step 2

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

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

Step 3

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

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

Step 4

Combine the numerators over the common denominator.

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

Step 5

Simplify the numerator.

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Step 5.1

Multiply $- 1$ by $3$ .

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

Step 5.2

Subtract $3$ from $1$ .

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

Step 6

Differentiate.

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Step 6.1

Move the negative in front of the fraction.

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

Step 6.2

Combine fractions.

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Step 6.2.1

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

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

Step 6.2.2

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

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

Step 6.3

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

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

Step 6.4

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

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

Step 7

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

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Step 7.1

To apply the Chain Rule, set $u_{2}$ as $\tan (2 x)$ .

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

Step 7.2

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

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

Step 7.3

Replace all occurrences of $u_{2}$ with $\tan (2 x)$ .

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

Step 8

Multiply $3$ by $2$ .

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

Step 9

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

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Step 9.1

To apply the Chain Rule, set $u_{3}$ as $2 x$ .

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

Step 9.2

The derivative of $\tan (u_{3})$ with respect to $u_{3}$ is $\sec^{2} (u_{3})$ .

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

Step 9.3

Replace all occurrences of $u_{3}$ with $2 x$ .

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

Step 10

Differentiate.

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Step 10.1

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

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

Step 10.2

Multiply $2$ by $6$ .

$\frac{1}{3 {(2 \tan^{3} (2 x) - 1)}^{\frac{2}{3}}} (12 \tan^{2} (2 x) \sec^{2} (2 x) \frac{d}{d x} [x] + \frac{d}{d x} [- 1])$

Step 10.3

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

$\frac{1}{3 {(2 \tan^{3} (2 x) - 1)}^{\frac{2}{3}}} (12 \tan^{2} (2 x) \sec^{2} (2 x) \cdot 1 + \frac{d}{d x} [- 1])$

Step 10.4

Multiply $12$ by $1$ .

$\frac{1}{3 {(2 \tan^{3} (2 x) - 1)}^{\frac{2}{3}}} (12 \tan^{2} (2 x) \sec^{2} (2 x) + \frac{d}{d x} [- 1])$

Step 10.5

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

$\frac{1}{3 {(2 \tan^{3} (2 x) - 1)}^{\frac{2}{3}}} (12 \tan^{2} (2 x) \sec^{2} (2 x) + 0)$

Step 10.6

Simplify terms.

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Step 10.6.1

Add $12 \tan^{2} (2 x) \sec^{2} (2 x)$ and $0$ .

$\frac{1}{3 {(2 \tan^{3} (2 x) - 1)}^{\frac{2}{3}}} (12 \tan^{2} (2 x) \sec^{2} (2 x))$

Step 10.6.2

Combine $12$ and $\frac{1}{3 {(2 \tan^{3} (2 x) - 1)}^{\frac{2}{3}}}$ .

$\frac{12}{3 {(2 \tan^{3} (2 x) - 1)}^{\frac{2}{3}}} (\tan^{2} (2 x) \sec^{2} (2 x))$

Step 10.6.3

Combine $\tan^{2} (2 x)$ and $\frac{12}{3 {(2 \tan^{3} (2 x) - 1)}^{\frac{2}{3}}}$ .

$\frac{\tan^{2} (2 x) \cdot 12}{3 {(2 \tan^{3} (2 x) - 1)}^{\frac{2}{3}}} \sec^{2} (2 x)$

Step 10.6.4

Combine $\frac{\tan^{2} (2 x) \cdot 12}{3 {(2 \tan^{3} (2 x) - 1)}^{\frac{2}{3}}}$ and $\sec^{2} (2 x)$ .

$\frac{\tan^{2} (2 x) \cdot 12 \sec^{2} (2 x)}{3 {(2 \tan^{3} (2 x) - 1)}^{\frac{2}{3}}}$

Step 10.6.5

Move $12$ to the left of $\tan^{2} (2 x)$ .

$\frac{12 \cdot \tan^{2} (2 x) \sec^{2} (2 x)}{3 {(2 \tan^{3} (2 x) - 1)}^{\frac{2}{3}}}$

Step 10.6.6

Factor $3$ out of $12 \tan^{2} (2 x) \sec^{2} (2 x)$ .

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

Step 11

Cancel the common factors.

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Step 11.1

Factor $3$ out of $3 {(2 \tan^{3} (2 x) - 1)}^{\frac{2}{3}}$ .

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

Step 11.2

Cancel the common factor.

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

Step 11.3

Rewrite the expression.

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

Step 12

Reorder terms.

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