Question

$If{x}^y=e^{x-y}then$

• A. $\frac{dy}{dx}does{n}^{\prime }texistatx=1$
• B. $\frac{dy}{dx}=0whenx=1$
• C. $\frac{dy}{dx}=\frac{1}{2}whenx=e$
• D. None of these

Answer 1

The correct option is A $\frac{dy}{dx}does{n}^{\prime }texistatx=1$

given,

$x^y=e^{x-y}$

let

$x^y=h$

$y=lo{g}_x^h=lo{g}_{e}h$

$\Rightarrow \frac{dy}{dx}=\frac{lo{g}_{e}x\left(\frac{1}{h}\right)\left(\frac{dh}{dx}\right)-lo{g}_{e}h\left(\frac{1}{x}\right)}{(lo{g}_{e}x{)}^2}$

$\therefore \frac{dh}{dx}\left[\frac{lo{g}_{e}x}{x^y}\right]=(lo{g}_{e}x{)}^2\frac{dy}{dx}+lo{g}_{e}x(\frac{y}{x})$

$\therefore \frac{dh}{dx}=\left[\right(lo{g}_{e}x{)}^2\frac{dy}{dx}+lo{g}_{e}x\left(\frac{y}{x}\right)]\left[\frac{x^y}{lo{g}_{e}x}\right]$

$\frac{d(x{)}^y}{dx}=\frac{d(e{)}^{x-y}}{dx}$

$\left[\right(lo{g}_{e}x{)}^2\frac{dy}{dx}+lo{g}_{e}x\left(\frac{y}{x}\right)]\left[\frac{x^y}{lo{g}_{e}x}\right]=(e{)}^{x-y}(1-\frac{dy}{dx})$

at $x=1$, above expression is not defined as$\frac{dy}{dx}$ itself is not defined .

In $\frac{dh}{dx}$ to cancel out out $logx,x$ can't be $1.$

if $x=1,$ $\frac{dh}{dx}$ is not defined

$\therefore \frac{dy}{dx}$ is also undefined .