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If $f\left(x^{5}\right)=5 x^{3}$, then $f^{\prime}(x)$ is equal to
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Verified Answer
The correct answer is:
$\frac{3}{\sqrt[5]{\mathrm{x}^{2}}}$
Given, $\mathrm{f}\left(\mathrm{x}^{5}\right)=5 \mathrm{x}^{3}$
Let $\quad \mathrm{x}^{5}=\mathrm{y} \Rightarrow \mathrm{x}^{3}=\mathrm{y}^{3 / 5}$
$\therefore \quad \mathrm{f}(\mathrm{y})=5 \mathrm{y}^{3 / 5}$
or $\quad \mathrm{f}(\mathrm{x})=5 \mathrm{x}^{3 / 5}$
On differentiating w.r.t. x, we get
$$
\begin{aligned}
f^{\prime}(x) &=5 \cdot \frac{3}{5} x^{-2} \\
&=\frac{3}{\sqrt[5]{x^{2}}}
\end{aligned}
$$
Let $\quad \mathrm{x}^{5}=\mathrm{y} \Rightarrow \mathrm{x}^{3}=\mathrm{y}^{3 / 5}$
$\therefore \quad \mathrm{f}(\mathrm{y})=5 \mathrm{y}^{3 / 5}$
or $\quad \mathrm{f}(\mathrm{x})=5 \mathrm{x}^{3 / 5}$
On differentiating w.r.t. x, we get
$$
\begin{aligned}
f^{\prime}(x) &=5 \cdot \frac{3}{5} x^{-2} \\
&=\frac{3}{\sqrt[5]{x^{2}}}
\end{aligned}
$$
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