Question

If $x=e^{y+e^{y+......\infty }},x>0,$ then $\frac{dy}{dx}$ is equal to

• A. $\frac{1-x}{x}$
• B. $\frac{1}{x}$
• C. $\frac{x}{1+x}$
• D. $\frac{1+x}{x}$

Answer 1

Answer: (A)

Given, $x=e^{y+x}$ $\Rightarrow$ $logx=x+y$ On differentiating w.r.t. $x,$ we get $\frac{1}{x}=1+\frac{dy}{dx}$ $\Rightarrow$ $\frac{dy}{dx}=\frac{1-x}{x}$