A) \[-\frac{y}{x}\]
B) \[-\frac{x}{y}\]
C) \[1+\log \left( \frac{x}{y} \right)\]
D) \[\frac{1+\log x}{1+\log y}\]
Correct Answer: D
Solution :
Given, \[{{x}^{x}}={{y}^{y}}\] Taking log on both sides, we get \[x\,\log \,x=y\,\log y\] Differentiating w.r.t. y, we get \[y.\frac{1}{y}.\frac{dy}{dx}+\log y\frac{dy}{dx}=x\frac{1}{x}+\log x\] \[\Rightarrow \] \[\frac{dy}{dx}(1+\log y)=1+\log x\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1+\log x}{1+\log y}\]
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