Question:
If \( x^y = e^{x - y} \), then \( \frac{dy}{dx} \) is equal to
If \( x^y = e^{x - y} \), then \( \frac{dy}{dx} \) is equal to
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If $x$, then $dy/dx$ is equal to
Updated On: Apr 15, 2026
- $\frac{log~x}{(1+log~x)^{2}}$
- $\frac{x-y}{(1+log~x)}$
- $\frac{x-y}{(1+log~x)^{2}}$
- $\frac{1}{(1+log~x)}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Take the natural logarithm on both sides: $y~log~x = (x-y)~log~e = x - y$.
Step 2: Analysis
Express $y$ in terms of $x$: $y(1 + log~x) = x \Rightarrow y = \frac{x}{1 + log~x}$.
Step 3: Evaluation
Differentiate using the quotient rule: $\frac{dy}{dx} = \frac{(1+log~x)\cdot 1 - x \cdot (1/x)}{(1+log~x)^2}$.
Step 4: Conclusion
Simplify: $\frac{dy}{dx} = \frac{1 + log~x - 1}{(1+log~x)^2} = \frac{log~x}{(1+log~x)^2}$.
Final Answer: (a)
Take the natural logarithm on both sides: $y~log~x = (x-y)~log~e = x - y$.
Step 2: Analysis
Express $y$ in terms of $x$: $y(1 + log~x) = x \Rightarrow y = \frac{x}{1 + log~x}$.
Step 3: Evaluation
Differentiate using the quotient rule: $\frac{dy}{dx} = \frac{(1+log~x)\cdot 1 - x \cdot (1/x)}{(1+log~x)^2}$.
Step 4: Conclusion
Simplify: $\frac{dy}{dx} = \frac{1 + log~x - 1}{(1+log~x)^2} = \frac{log~x}{(1+log~x)^2}$.
Final Answer: (a)
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