Question

What is the derivative of `sin(x^2y^2)`?

Answer 1

Answer 1
If you want the partial derivatives of `f(x,y)=sin(x^2y^2)`, they are:

`f_x(x,y)=2xy^2cos(x^2y^2)` and

`f_y(x,y)=2x^2ycos(x^2y^2)`.

Answer 2
If we are considering `y` to be a function of `x` and looking for `d/(dx)(sin(x^2y^2))`, the the answer is:

`d/(dx)(sin(x^2y^2))=[2xy^2+2x^2y (dy)/(dx)]cos(x^2y^2)`

Find this using implicit differentiation (the chain rule) and the product rule.

`d/(dx)(sin(x^2y^2))=[cos(x^2y^2)]*d/(dx)(x^2y^2)`

`==[cos(x^2y^2)]*[2xy^2+x^2 2y (dy)/(dx)]`

`=[2xy^2+2x^2y (dy)/(dx)]cos(x^2y^2)`