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

Answer

$ y=\sqrt[5]{\left(3 x^2+8 x+5\right)^4}$
$y=\left(3 x^2+8 x+5\right)^{\frac{4}{5}} $
Differentiating both sides w.r.t. $x$, we get
$ \frac{ d y}{ d x}=\frac{ d }{ d x}\left[\left(3 x^2+8 x+5\right)^{\frac{4}{5}}\right]$
$=\frac{4}{5}\left(3 x^2+8 x+5\right)^{-\frac{1}{5}} \cdot \frac{ d }{ d x}\left(3 x^2+8 x+5\right)$
$=\frac{4}{5}\left(3 x^2+8 x+5\right)^{-\frac{1}{5}} \cdot[3(2 x)+8+0]$
$\therefore \frac{ d y}{ d x}=\frac{4}{5}\left(3 x^2+8 x+5\right)^{\frac{1}{5}} \cdot(6 x+8) $

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