Question

y=sin( x^2 y^2 )
y'=?

Answer 1

apply chain rule

Answer 2

Any ideas?

Answer 3

@jiteshmeghwal9 could you please explain more?

Answer 4

assume x^2y^2=u
sin(x^2y^2)=sinu

\[\frac{d (sinu)}{du} * \frac{du}{dx}\]

Answer 5

\[\frac{dy}{dx}=\frac{d(sinu)}{du}*\frac{d(x^2y^2)}{dx}\]