Calculus Examples

$f (t) = {(9 t + 2)}^{\frac{2}{3}}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{\frac{2}{3}}$ and $g (t) = 9 t + 2$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $9 t + 2$ .

$\frac{d}{d u} [u^{\frac{2}{3}}] \frac{d}{d t} [9 t + 2]$

Step 1.2

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

$\frac{2}{3} u^{\frac{2}{3} - 1} \frac{d}{d t} [9 t + 2]$

Step 1.3

Replace all occurrences of $u$ with $9 t + 2$ .

$\frac{2}{3} {(9 t + 2)}^{\frac{2}{3} - 1} \frac{d}{d t} [9 t + 2]$

Step 2

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

$\frac{2}{3} {(9 t + 2)}^{\frac{2}{3} - 1 \cdot \frac{3}{3}} \frac{d}{d t} [9 t + 2]$

Step 3

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

$\frac{2}{3} {(9 t + 2)}^{\frac{2}{3} + \frac{- 1 \cdot 3}{3}} \frac{d}{d t} [9 t + 2]$

Step 4

Combine the numerators over the common denominator.

$\frac{2}{3} {(9 t + 2)}^{\frac{2 - 1 \cdot 3}{3}} \frac{d}{d t} [9 t + 2]$

Step 5

Simplify the numerator.

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Step 5.1

Multiply $- 1$ by $3$ .

$\frac{2}{3} {(9 t + 2)}^{\frac{2 - 3}{3}} \frac{d}{d t} [9 t + 2]$

Step 5.2

Subtract $3$ from $2$ .

$\frac{2}{3} {(9 t + 2)}^{\frac{- 1}{3}} \frac{d}{d t} [9 t + 2]$

Step 6

Combine fractions.

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Step 6.1

Move the negative in front of the fraction.

$\frac{2}{3} {(9 t + 2)}^{- \frac{1}{3}} \frac{d}{d t} [9 t + 2]$

Step 6.2

Combine $\frac{2}{3}$ and ${(9 t + 2)}^{- \frac{1}{3}}$ .

$\frac{2 {(9 t + 2)}^{- \frac{1}{3}}}{3} \frac{d}{d t} [9 t + 2]$

Step 6.3

Move ${(9 t + 2)}^{- \frac{1}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{2}{3 {(9 t + 2)}^{\frac{1}{3}}} \frac{d}{d t} [9 t + 2]$

Step 7

By the Sum Rule, the derivative of $9 t + 2$ with respect to $t$ is $\frac{d}{d t} [9 t] + \frac{d}{d t} [2]$ .

$\frac{2}{3 {(9 t + 2)}^{\frac{1}{3}}} (\frac{d}{d t} [9 t] + \frac{d}{d t} [2])$

Step 8

Since $9$ is constant with respect to $t$ , the derivative of $9 t$ with respect to $t$ is $9 \frac{d}{d t} [t]$ .

$\frac{2}{3 {(9 t + 2)}^{\frac{1}{3}}} (9 \frac{d}{d t} [t] + \frac{d}{d t} [2])$

Step 9

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

$\frac{2}{3 {(9 t + 2)}^{\frac{1}{3}}} (9 \cdot 1 + \frac{d}{d t} [2])$

Step 10

Multiply $9$ by $1$ .

$\frac{2}{3 {(9 t + 2)}^{\frac{1}{3}}} (9 + \frac{d}{d t} [2])$

Step 11

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

$\frac{2}{3 {(9 t + 2)}^{\frac{1}{3}}} (9 + 0)$

Step 12

Simplify terms.

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Step 12.1

Add $9$ and $0$ .

$\frac{2}{3 {(9 t + 2)}^{\frac{1}{3}}} \cdot 9$

Step 12.2

Combine $\frac{2}{3 {(9 t + 2)}^{\frac{1}{3}}}$ and $9$ .

$\frac{2 \cdot 9}{3 {(9 t + 2)}^{\frac{1}{3}}}$

Step 12.3

Multiply $2$ by $9$ .

$\frac{18}{3 {(9 t + 2)}^{\frac{1}{3}}}$

Step 12.4

Factor $3$ out of $18$ .

$\frac{3 \cdot 6}{3 {(9 t + 2)}^{\frac{1}{3}}}$

Step 13

Cancel the common factors.

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Step 13.1

Factor $3$ out of $3 {(9 t + 2)}^{\frac{1}{3}}$ .

$\frac{3 \cdot 6}{3 ({(9 t + 2)}^{\frac{1}{3}})}$

Step 13.2

Cancel the common factor.

$\frac{3 \cdot 6}{3 {(9 t + 2)}^{\frac{1}{3}}}$

Step 13.3

Rewrite the expression.

$\frac{6}{{(9 t + 2)}^{\frac{1}{3}}}$