A) \[\frac{\log x}{{{(1+\log x)}^{2}}}\]
B) \[\frac{y-x}{x(1+\log x)}\]
C) \[\frac{x-y}{1+\log \,\,x}\]
D) \[\frac{y-x}{x(1+log\,\,x)}\]
Correct Answer: A
Solution :
\[{{x}^{y}}={{e}^{x-y}}\] Taking log both the sides, \[y\text{ }log\text{ }x=(x-y)log\text{ }e\] \[\Rightarrow \] \[y\,log\,x=x-y\] ...(i) On differentiating w.r.t.\['x',\] \[y.\frac{1}{x}+\log x\frac{dy}{dx}=1-\frac{dy}{dx}\] \[\Rightarrow \] \[\frac{dy}{dx}(\log x+1)=1-\frac{y}{x}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{x-y}{x(\log x+1)}\] \[=\frac{y\log x}{(y\log x+y)(\log x+1)}\] Putting value in Eq. (i), \[=\frac{y\log x}{y(\log x+1)(\log x+1)}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{\log x}{{{(1+\log x)}^{2}}}\]
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