Question

If $ {{x}^{y}}={{e}^{x-y}} $ , then $ \frac{dy}{dx} $ is:

• 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}}} $

Answer 1

Answer: (D)

$ {{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}}} $