Question

If $y=f\left(x\right)\text{satisfies}2e^{x{y}^2}+ycos\left(x^2\right)=4$ then value of $\left|\frac{f^{\prime }\left(0\right)}{5}\right|$ is

Answer 1

at x = 0, y = 2
$2e^{x{y}^2}(y^2+2xy{y}^{\prime })+\left(cos{x}^2\right)y^{\prime }-ysin\left(x^2\right).2x=0$
at $x=0$ and $y=2$
$2(4+0)+y^{\prime }\left(0\right)=0$
$y^{\prime }\left(0\right)=-8\Rightarrow f^{\prime }\left(0\right)=-8$
$\left|\frac{f^{\prime }\left(0\right)}{5}\right|=\left|\frac{-8}{5}\right|=1.6$