What is tan^3 (xy^2+y) = x?

Question

Calculus: Implicit Differentiation

Answer 1

If we have an implicit function f(x,y)=0, then the complete differential is 

f'_xdx+f'_ydy=0, (f'_x=df/dx, f'_y=df/dy).

Hence y'=dy/dx=-f'_x/f'_y

Now f(x,y)=tan³(xy²+y) - x

df/dx= -1+3 y² sec(y + x y²)² tan(y + x y²)²

df/dy=3 (1+2xy) sec(y + x y²)² tan(y + x y²)²

If you simplify y'=-f'x/f'y, you get

 

y'=(-3 y² + cos(y(1 + x y))² cot(y (1 + x y))²)/(3(1 + 2 x y)).

 

Answer 2

This ought to be stated more clearly as a question.

Possibly you mean: if tan^3 (xy^2+y) = x, find dy/dx

 

I would try to help you, IF THAT is the intended question.