Find d y d x if, : y= [3] a^2+x^2

Question

Find $\frac{d y}{d x}$ if, :
$
y=\sqrt[3]{a^2+x^2}
$

✓

Answer

Given : $y=\sqrt[3]{a^2+x^2}$
Let $u=a^2+x^2$
Then $y=\sqrt[3]{u}$
$
\begin{aligned}
\therefore \frac{d y}{d u}=\frac{d}{d u}\left(u^{\frac{1}{3}}\right) & =\frac{1}{3} u^{-\frac{2}{3}} \\
& =\frac{1}{3}\left(a^2+x^2\right)^{-\frac{2}{3}}
\end{aligned}
$
$
\text { and } \begin{aligned}
\frac{d u}{d x} & =\frac{d}{d x}\left(a^2+x^2\right) \\
& =0+2 x=2 x
\end{aligned}
$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\
& =\frac{1}{3}\left(a^2+x^2\right)^{-\frac{2}{3}} \cdot 2 x \\
& =\frac{2 x}{3}\left(a^2+x^2\right)^{-\frac{2}{3}}
\end{aligned}
$

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