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Q.
The curve $y-e^{xy} + x = 0$ has a vertical tangent at the point
Application of Derivatives
Solution:
$y - e^{xy}+x = 0$
$\Rightarrow \frac{dy}{dx} - e^{xy}\left[x \frac{dy}{dx}+y\cdot1\right]+1=0$
$\Rightarrow \left(1-xe^{xy}\right) \frac{dy}{dx}=ye^{xy}-1$
$\Rightarrow \frac{dy}{dx} = \frac{ye^{xy}-1}{1-xe^{xy}}$
$\Rightarrow \frac{dx}{dy} = \frac{1-xe^{xy}}{ye^{xy}-1}$
For a vertical tangent, $\frac{dx}{dy}=0$
which holds at $\left(1, 0\right)$. This pt. also lies on the curve.