KCET · Maths · Differentiation
If \(f\left(x^{5}\right)=5 x^{3}\), then \(f^{\prime}(x)\) is equal to
- A \(\frac{3}{\sqrt[5]{\mathrm{x}^{2}}}\)
- B \(\frac{3}{\sqrt[5]{x}}\)
- C \(\frac{3}{x}\)
- D \(\sqrt[5]{x}\)
Answer & Solution
Correct Answer
(A) \(\frac{3}{\sqrt[5]{\mathrm{x}^{2}}}\)
Step-by-step Solution
Detailed explanation
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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