Find the Derivative - d/dx cube root of 2x+3

$\sqrt[3]{2 x + 3}$

Step 1

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

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

Step 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 $g (x) = 2 x + 3$ .

To apply the Chain Rule, set $u$ as $2 x + 3$ .

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

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

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

Replace all occurrences of $u$ with $2 x + 3$ .

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

Step 3

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

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

Step 4

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

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

Step 5

Combine the numerators over the common denominator.

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

Step 6

Simplify the numerator.

Multiply $- 1$ by $3$ .

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

Subtract $3$ from $1$ .

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

Step 7

Combine fractions.

Move the negative in front of the fraction.

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

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

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

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

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

Step 8

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

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

Step 9

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

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

Step 10

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

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

Step 11

Multiply $2$ by $1$ .

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

Step 12

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

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

Step 13

Combine fractions.

Add $2$ and $0$ .

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

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

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

Find the Derivative - d/dx cube root of 2x+3

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