Question

The curve given by the equation $y-e^{xy}+x=0$ has a vertical tangent at the point

• A. $(0,1)$
• B. $(1,1)$
• C. $(-1,1)$
• D. $(1,0)$

Answer 1

The correct option is D $(1,0)$

$y-e^{xy}+x=0$
$\therefore \frac{dy}{dx}-e^{xy}(x\frac{dy}{dx}+y)+1=0$
$\Rightarrow \frac{dy}{dx}=\frac{-1+y{e}^{xy}}{1-x{e}^{xy}}$
$\therefore$ if the tangent is vertical at $(x,y),x{e}^{xy}=1$ 1 which is possible only if $x=1,y=0.$