Find the derivative of (x^3)/((x+3)^(1/3))

\frac{d}{dx}\left(\frac{x^3}{\left(x+3\right)^{\frac{1}{3}}}\right)

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Answer

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

Step by step solution

Problem

$\frac{d}{dx}\left(\frac{x^3}{\left(x+3\right)^{\frac{1}{3}}}\right)$
1

Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{\sqrt[3]{3+x}\cdot\frac{d}{dx}\left(x^3\right)-x^3\frac{d}{dx}\left(\sqrt[3]{3+x}\right)}{\left(\sqrt[3]{3+x}\right)^2}$
2

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

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

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

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

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{3\sqrt[3]{3+x}x^{2}-1\cdot \frac{1}{3}x^3\left(3+x\right)^{-\frac{2}{3}}\left(\frac{d}{dx}\left(3\right)+\frac{d}{dx}\left(x\right)\right)}{\left(\sqrt[3]{3+x}\right)^2}$
5

The derivative of the constant function is equal to zero

$\frac{3\sqrt[3]{3+x}x^{2}-1\cdot \frac{1}{3}x^3\left(3+x\right)^{-\frac{2}{3}}\left(0+\frac{d}{dx}\left(x\right)\right)}{\left(\sqrt[3]{3+x}\right)^2}$
6

The derivative of the linear function is equal to $1$

$\frac{3\sqrt[3]{3+x}x^{2}-1\cdot \left(0+1\right)\cdot \frac{1}{3}x^3\left(3+x\right)^{-\frac{2}{3}}}{\left(\sqrt[3]{3+x}\right)^2}$
7

Add the values $1$ and $0$

$\frac{3\sqrt[3]{3+x}x^{2}-1\cdot 1\cdot \frac{1}{3}x^3\left(3+x\right)^{-\frac{2}{3}}}{\left(\sqrt[3]{3+x}\right)^2}$
8

Multiply $-1$ times $\frac{1}{3}$

$\frac{3\sqrt[3]{3+x}x^{2}-\frac{1}{3}x^3\left(3+x\right)^{-\frac{2}{3}}}{\left(\sqrt[3]{3+x}\right)^2}$
9

Applying the power of a power property

$\frac{3\sqrt[3]{3+x}x^{2}-\frac{1}{3}x^3\left(3+x\right)^{-\frac{2}{3}}}{\sqrt[3]{\left(3+x\right)^{2}}}$
10

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$\frac{3\sqrt[3]{3+x}x^{2}-\frac{1}{3}x^3\frac{1}{\sqrt[3]{\left(3+x\right)^{2}}}}{\sqrt[3]{\left(3+x\right)^{2}}}$
11

Apply the formula: $a\frac{1}{x}$$=\frac{a}{x}$, where $a=-\frac{1}{3}$ and $x=\sqrt[3]{\left(3+x\right)^{2}}$

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

Answer

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

Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.27 seconds

Views:

97