Question

State True or False.

If $\frac{e^y}{e^x}=xy$, then $y^{\prime }=\frac{2-\log x}{{\left(1-\log x\right)}^2}$

• A. True
• B. False

Answer 1

The correct option is A True

$\frac{e^y}{e^x}=xy$

$\Rightarrow e^{y-x}=xy$

$\Rightarrow (y-x)\log e=\log xy$

$\Rightarrow y-x=\log x+\log y$

$\Rightarrow \frac{dy}{dx}-1=\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}(1-\frac{1}{y})=1+\frac{1}{x}$

$\Rightarrow \frac{dy}{dx}\left(\frac{y-1}{y}\right)=\frac{x+1}{x}$

$\Rightarrow \frac{dy}{dx}=\frac{y(x+1)}{x(y-1)}$

Given statement is false.