KCET · Maths · Differentiation
If \( x^{y}=e^{x-y} \) then \( \frac{d y}{d x} \) is equal to
- A \( \frac{\log x}{\log (x-y)} \)
- B \( \frac{e^{x}}{x^{x-y}} \)
- C \( \frac{\log x}{(1+\log x)^{2}} \)
- D \( \frac{1}{y}-\frac{1}{x-y} \)
Answer & Solution
Correct Answer
(C) \( \frac{\log x}{(1+\log x)^{2}} \)
Step-by-step Solution
Detailed explanation
Given that, \( x^{y}=e^{x-y} \)
Taking log both sides, we get
\[
\begin{array}{l}
y \log x=x-y \\
y(1+\log x)=x \\
\Rightarrow y=\frac{x}{(1+\log x)}
\end{array}
\]
\[
\begin{array}{l}
\text { So, } \frac{d y}{d x}=\frac{(1+\log x)-x\left(\frac{1}{x}\right)}{(1+\log x)^{2}} \\
=\frac{\log x}{(1+\log x)^{2}}
\end{array}
\]
Taking log both sides, we get
\[
\begin{array}{l}
y \log x=x-y \\
y(1+\log x)=x \\
\Rightarrow y=\frac{x}{(1+\log x)}
\end{array}
\]
\[
\begin{array}{l}
\text { So, } \frac{d y}{d x}=\frac{(1+\log x)-x\left(\frac{1}{x}\right)}{(1+\log x)^{2}} \\
=\frac{\log x}{(1+\log x)^{2}}
\end{array}
\]
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