If y = [5] (3 x^2+8 x+5 )^4 , find d y d x - Vidyadip

Question

If $y =\sqrt[5]{\left(3 x^2+8 x+5\right)^4}$, find $\frac{d y}{d x}$

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Answer

$
\text { Given : } y=\sqrt[5]{\left(3 x^2+8 x+5\right)^4}
$
Let $u=3 x^2+8 x+5$
Then $y=\sqrt[5]{u^4}=u^{\frac{4}{5}}$
$
\begin{aligned}
\therefore \frac{d y}{d u} & =\frac{d}{d u}\left(u^{\frac{4}{5}}\right)=\frac{4}{5} u^{\frac{4}{5}-1} \\
& =\frac{4}{5}\left(3 x^2+8 x+5\right)^{-\frac{1}{5}}
\end{aligned}
$
$
\text { and } \begin{aligned}
\frac{d u}{d x} & =\frac{d}{d x}\left(3 x^2+8 x+5\right) \\
& =3 \frac{d}{d x}\left(x^2\right)+8 \frac{d}{d x}(x)+\frac{d}{d x}(5) \\
& =3 \times 2 x+8 \times 1+0=6 x+8
\end{aligned}
$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\
& =\frac{4}{5}\left(3 x^2+8 x+5\right)^{-\frac{1}{5}} \cdot(6 x+8) .
\end{aligned}
$

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