A) \[\frac{1+x}{1+\log x}\]
B) \[\frac{1-\log x}{1+\log x}\]
C) not defined
D) \[\frac{\log x}{{{(1+\log x)}^{2}}}\]
Correct Answer: D
Solution :
\[{{x}^{y}}={{e}^{x-y}}\] Taking log on both sides, we get \[y\log x=x-y\] \[\Rightarrow \] \[y(\log x+1)=x\] \[\Rightarrow \] \[y=\frac{x}{1+\log x}\] On differentiating w.r.t\[x,\] we get \[\Rightarrow \] \[\frac{dy}{dx}=\frac{(1+\log x)-x\times \frac{1}{x}}{{{(1+\log x)}^{2}}}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{\log x}{{{(1+\log x)}^{2}}}\]
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