Mathematics - Application of Derivative Question with Solution | TestHub

$\begin{array}{rl}& y-e^{xy}+x=0\\ & \therefore \frac{dy}{dx}-e^{xy}\left(y+x\frac{dy}{dx}\right)+1=0\\ & \text{i.e., }\frac{dy}{dx}-y(x+y)-x(x+y)\frac{dy}{dx}+1=0\\ & \text{i.e., }[1-x(x+y)]\frac{dy}{dx}=y(x+y)-1\end{array}$ for the vertical tangents $1-x(x+y)=0$ i.e., $y=\frac{1-x^2}{x}$ $\therefore \mathrm{x}=1$ and $\mathrm{y}=0$