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MHT CET · Maths · Differentiation

For \(x>1\), if \((2 x)^{2 y}=4 e^{2 x-2 y}\), then \(\left(1+\log _e 2 x\right)^2 \frac{\mathrm{d} y}{\mathrm{~d} x}\) is equal to

  1. A \(x \log _{\mathrm{e}} 2 x\)
  2. B \(\log _e 2 x\)
  3. C \(\frac{x \log _{\mathrm{e}} 2 x+\log _{\mathrm{e}} 2}{x}\)
  4. D \(\frac{x \log _{\mathrm{e}} 2 x-\log _{\mathrm{e}} 2}{x}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{x \log _{\mathrm{e}} 2 x-\log _{\mathrm{e}} 2}{x}\)

Step-by-step Solution

Detailed explanation

\((2 x)^{2 y}=4 \cdot e^{2 x-2 y}\)
\(\Rightarrow 2 y \log 2 x=\log 4+2 x-2 y\)
\(\Rightarrow y=\frac{x+\log 2}{1+\log 2 x}\)
\(\begin{aligned} & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{(1+\log 2 x)-(x+\log 2) \cdot \frac{1}{2 x} \cdot 2}{(1+\log 2 x)^2} \\ & \Rightarrow(1+\log 2 x)^2 \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x \log 2 x-\log 2}{x}\end{aligned}\)