Question

using implicit differentiation differentiate
tan^3 xy^2+y=x

Answer 1

\[\tan^{3} x(y^{2}) +y = x\]
is that correct?

Answer 2

I will assume it is
\[\rightarrow 3y^{2}\tan^{2}x \sec^{2}x + 2y \tan^{3}x \frac{dy}{dx} + \frac{dy}{dx} = 1\]
solve for dy/dx
\[\frac{dy}{dx} = \frac{1 - 3y^{2} \tan^{2} x \sec^{2}x}{1+2y \tan^{3}x}\]

Answer 3

3tan^2x sec^2x y^2 + tan^3x (2yy') + y' = 1
tan^3x (2yy') + y' = 1 - 3tan^2x sec^2x y^2
(2tan^3xy + 1) y' = 1 - 3tan^2x sec^2x y^2
=> y' = {1 - 3tan^2x sec^2x y^2 / (2tan^3xy + 1)

Answer 4

@chlorophyll i forgot to introduce brackets at (xy^2+y)