Mathematics - Differentiation Question with Solution | TestHub

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