MHT CET · Maths · Differentiation
Derivative of \(\mathrm{e}^x\) w.r.t. \(\sqrt{x}\) is
- A \(\sqrt{x} e^x\)
- B \(-2 \sqrt{x}\)
- C \(2 \sqrt{x} \mathrm{e}^x\)
- D \(\frac{1}{2} \sqrt{x} \mathrm{e}^x\)
Answer & Solution
Correct Answer
(C) \(2 \sqrt{x} \mathrm{e}^x\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{u}=\mathrm{e}^x\) and \(\mathrm{v}=\sqrt{x}\) Differentiating u and v w.r.t. \(x\), we get
\(\begin{aligned}
& \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{e}^x, \frac{\mathrm{dv}}{\mathrm{~d} x}=\frac{1}{2 \sqrt{x}} \\
\therefore \quad & \frac{\mathrm{du}}{\mathrm{dv}}=\frac{\frac{\mathrm{du}}{\mathrm{dx}}}{\frac{\mathrm{dv}}{\mathrm{dx}}}=\frac{\mathrm{e}^x}{\frac{1}{2 \sqrt{x}}}=2 \mathrm{e}^x \sqrt{x}
\end{aligned}\)
\(\begin{aligned}
& \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{e}^x, \frac{\mathrm{dv}}{\mathrm{~d} x}=\frac{1}{2 \sqrt{x}} \\
\therefore \quad & \frac{\mathrm{du}}{\mathrm{dv}}=\frac{\frac{\mathrm{du}}{\mathrm{dx}}}{\frac{\mathrm{dv}}{\mathrm{dx}}}=\frac{\mathrm{e}^x}{\frac{1}{2 \sqrt{x}}}=2 \mathrm{e}^x \sqrt{x}
\end{aligned}\)
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