Step-by-step Solution

Find the derivative using the quotient rule $\frac{d}{dx}\left(\frac{x^3}{\sqrt[3]{x+3}}\right)$

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

Step-by-step explanation

Problem to solve:

$\frac{d}{dx}\left(\frac{x^3}{\left(x+3\right)^{\frac{1}{3}}}\right)$
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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

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

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

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

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