MHT CET · Maths · Differentiation
If \(x^y=e^{x-y}\), then \(\frac{d y}{d x}=\)
- A \(\frac{\log x}{1+\log x}\)
- B \(\frac{\log x}{x(1+\log x)^2}\)
- C \(\frac{\log x}{(1+\log x)^2}\)
- D \(\frac{x \log x}{(1+\log x)^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\log x}{(1+\log x)^2}\)
Step-by-step Solution
Detailed explanation
\(x^y=e^{x-y}\)
\(\begin{aligned} & \Rightarrow \frac{d y}{d x} \log x+y \cdot \frac{1}{x}=1-\frac{d y}{d x} \\ & \Rightarrow \frac{d y}{d x}=\frac{x-y}{x(1+\log x)} \\ & \Rightarrow \frac{d y}{d x}=\frac{\log x}{(1+\log x)^2} \left[\text { Putting } y=\frac{x}{1+\log x} \text { from (i)] }\right.\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{d y}{d x} \log x+y \cdot \frac{1}{x}=1-\frac{d y}{d x} \\ & \Rightarrow \frac{d y}{d x}=\frac{x-y}{x(1+\log x)} \\ & \Rightarrow \frac{d y}{d x}=\frac{\log x}{(1+\log x)^2} \left[\text { Putting } y=\frac{x}{1+\log x} \text { from (i)] }\right.\end{aligned}\)
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