Question

How do you find `(dy)/(dx)` given `y^2+siny=sinx`?

Answer 1

`dy/dx=(cosx)/(2y+cosy)`

Explanation:

differentiate `color(blue)"implicitly with respect to x"`

`2y.dy/dx+cosy.dy/dx=cosx`

`rArrdy/dx(2y+cosy)=cosx`

`rArrdy/dx=(cosx)/(2y+cosy)`