A) \[\frac{1+x}{1+\log x}\]
B) \[\frac{1-\log x}{1+\log x}\]
C) not defined
D) \[\frac{logx}{(1+\log x)}\]
Correct Answer: D
Solution :
\[{{x}^{y}}={{e}^{x-y}}\] Taking the log of both sides \[y\text{ }log\text{ }x=x-y\] \[\Rightarrow \] \[y=\frac{x}{1+\log x}\] \[\frac{dy}{dx}=\frac{(1+\log x)-x\times \frac{1}{x}}{{{(1+\log x)}^{2}}}\] \[\frac{dy}{dx}=\frac{\log x}{{{(1+\log x)}^{2}}}\]You need to login to perform this action.
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