Question

If $y=y\left(x\right)$ and it follows the relation $2e^{x{y}^2}+y\cos \left(x^2\right)=4,$ then $|y^{\prime }\left(0\right)|$ is equal to

Answer 1

at $x=0\Rightarrow 2+ycos0=4$
$\Rightarrow y=2$
Differentiate it w.r.t. x
$\Rightarrow 2e^{x{y}^2}(y^2+2xy.y^{\prime })+y^{\prime }.cos{x}^2-2xysin\left(x^2\right)=0$
put $x=0,y=2$
$y^{\prime }\left(0\right)=-8$
$\Rightarrow |y^{\prime }\left(0\right)|=8$