Question

If $e^{x{y}^2}+y\cos x^2=5,$ then absolute value of $y^{\prime }\left(0\right)=$

Answer 1

We have,
$e^{x{y}^2}+y\cos x^2=5\cdots \left(1\right)$
Differentiating both sides w.r.t. $x,$ we get
$e^{x{y}^2}(y^2+2xy\frac{dy}{dx})+\frac{dy}{dx}\cos x^2-y\sin x^2\times 2x=0$
$\Rightarrow \frac{dy}{dx}(2xy\cdot e^{x{y}^2}+\cos x^2)=2xy\sin x^2-y^2\cdot e^{x{y}^2}$
$\Rightarrow \frac{dy}{dx}=\frac{2xy\sin x^2-y^2\cdot e^{x{y}^2}}{2xy\cdot e^{x{y}^2}+\cos x^2}$
Putting $x=0$ in equation $\left(1\right),$ we get
$e^0+y\cos \left(0\right)=5$
$\Rightarrow y=4$
$\Rightarrow y^{\prime }\left(0\right)=\frac{0-16}{1}=-16$
$\therefore |y^{\prime }\left(0\right)|=16$