Thank you for reporting, we will resolve it shortly
Q.
If $y=y(x)$ and it follows the relation $e^{x y^2}+y \cos \left(x^2\right)=5$ then $y^{\prime}(0)$ is equal to
Continuity and Differentiability
Solution:
We have $e ^{ xy }{ }^2+ y \cos \left( x ^2\right)=5$ .....(1)
Put $x =0$, we get
$1+ y =5 \Rightarrow y =4 \Rightarrow(0,4)$ lies on the given curve.
Now, differentiating (1) with respect to $x$, we get
$\Rightarrow e ^{ xy }\left[ x \cdot 2 y \cdot \frac{ dy }{ dx }+ y ^2\right]- y \cdot 2 x \cdot \sin \left( x ^2\right)+\cos \left( x ^2\right) \frac{ dy }{ dx }=0$
As $(0,4)$ satisfy it, we get
$\left.\left.16+\frac{ dy }{ dx }\right]_{(0,4)}=0 \Rightarrow \frac{ dy }{ dx }\right]_{(0,4)}=-16 $