A) \[\frac{\log x}{1+\log x}\]
B) \[\frac{\log x}{1-\log x}\]
C) \[\frac{\log x}{{{(1+\log x)}^{2}}}\]
D) \[\frac{y\log x}{x(1+\log x)}\]
E) \[\frac{1+\log x}{\log x}\]
Correct Answer: D
Solution :
\[{{x}^{y}}={{e}^{x-y}}\] \[\Rightarrow \] \[y\log x=(x-y)\] On differentiating w. r. t.\[x,\]we get \[\Rightarrow \]\[y\frac{1}{x}+\log x\frac{dy}{dx}=1-\frac{dy}{dx}\] \[\Rightarrow \]\[\frac{dy}{dx}(1+\log x)=1-\frac{y}{x}\] \[\Rightarrow \]\[\frac{dy}{dx}=\frac{x-y}{x(1+\log x)}=\frac{y\log x}{x(1+\log x)}\] \[(\because x-y=y\log x)\]
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