Question

How do you differentiate `3xy^2+cosy^2=2x^3+5`?

Answer 1

`dy/dx=(3(2x^2-y^2))/(2y(3x-siny^2))`

Explanation:

`d/dx(3xy^2+cosy^2)=d/dx(2x^3+5)`

`d/dx3xy^2+d/dxcosy^2=d/dx2x^3+d/dx5`

`[d/dx3xy^2]+d/dxcosy^2=d/dx2x^3+d/dx5`

`[3x2ydy/dx+y^2 3]+(-2ysiny^2)dy/dx=6x^2+0`
differentiate of `siny^2` and `sin^2y` is different answer

`6xydy/dx+3y^2 -(2ysiny^2) dy/dx=6x^2`

`dy/dx(6xy -2ysiny^2) =6x^2-3y^2`
`dy/dx=(6x^2-3y^2)/(6xy -2ysiny^2)=(3(2x^2-y^2))/(2y(3x-siny^2))`