A) \[\frac{\log x}{{{(1+\log x)}^{2}}}\]
B) \[\frac{x-y}{(1+\log x)}\]
C) \[\frac{x-y}{{{(1+\log x)}^{2}}}\]
D) \[\frac{1}{(1+\log x)}\]
Correct Answer: A
Solution :
We have,\[{{x}^{y}}={{e}^{x-y}}\] \[\Rightarrow \] \[y\log x=(x-y)\log e=x-y\] \[\Rightarrow \] \[y=\frac{x}{1+\log x}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{(1+\log x)\cdot 1-x\cdot \frac{1}{x}}{{{(1+\log x)}^{2}}}=\frac{\log x}{{{(1+\log x)}^{2}}}\]
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