Question

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

• A. $\frac{1+\log y}{y\log y}$
• B. $\frac{(1+\log y{)}^2}{y\log y}$
• C. $\frac{1+\log y}{(\log y{)}^2}$
• D. $\frac{(1+\log y{)}^2}{\log y}$

Answer 1

Answer: (D)

Here, $y^x=e^{y-x}$
Taking log on both sides, we get
$\log y^x=\log e^{y-x}$
$(\because \log a^b=b\log a$ and $\log e=1)$
$\Rightarrow x\log y=(y-x)\log e$
$\Rightarrow x\log y=y-x\dots (i)$
On differentiating w.r.t. $x$, we get
$\frac{d}{dx}(x\log y)=\frac{d}{dx}(y-x)$ (using product rule)
$\Rightarrow x\left(\frac{1}{y}\right)\frac{dy}{dx}+\log y(1)=\frac{dy}{dx}-1$
$\Rightarrow \frac{dy}{dx}\left(\frac{x}{y}-1\right)=-1-\log y$
$\Rightarrow \frac{dy}{dx}\left[\frac{y}{(1+\log y)y}-1\right]=-(1+\log y)$
$[\because$ from eq. $(i),x=\frac{y}{(1+\log y)}]$
$\Rightarrow \frac{dy}{dx}\left[\frac{1-1-\log y}{1+\log y}\right]=-(1+\log y)$
$\Rightarrow \frac{dy}{dx}=-\frac{(1+\log y{)}^2}{-\log y}$
$\Rightarrow \frac{dy}{dx}=\frac{(1+\log y{)}^2}{\log y}$