Question

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

• A. $(1,1)$
• B. at no point
• C. $(0,0)$
• D. $(1,0)$

Answer 1

Answer: (D)

$y-e^{xy}+x=0$
$\Rightarrow \frac{dy}{dx}-e^{xy}\left[x\frac{dy}{dx}+y\cdot 1\right]+1=0$
$\Rightarrow \left(1-x{e}^{xy}\right)\frac{dy}{dx}=y{e}^{xy}-1$
$\Rightarrow \frac{dy}{dx}=\frac{y{e}^{xy}-1}{1-x{e}^{xy}}$
$\Rightarrow \frac{dx}{dy}=\frac{1-x{e}^{xy}}{y{e}^{xy}-1}$
For a vertical tangent, $\frac{dx}{dy}=0$
which holds at $\left(1,0\right)$. This pt. also lies on the curve.