Question

How do you implicitly differentiate `e^(3x)/y^2=xsin(x-2y)`?

Answer 1

See below.

Explanation:

`frac{D_x e^(3x) y^2 - e^(3x) D_x y^2}{y^4} = 1sin(x - 2y) + x cos (x - 2y) (1 - 2y')`

`frac{3 e^(3x) y^2 - e^(3x) 2y'y}{y^4} = sin(x - 2y) + x cos (x - 2y) (1 - 2y')`

`y' = frac{y^4 sin(x - 2y) + xy^4 cos (x - 2y) - 3 e^(3x) y^2}{2xy^4 cos (x - 2y) - e^(3x) 2y}`