tan^3 (xy^2 + y) = x

Question

use implicit differentiation

Answer 1

We treat all variables when taking the implicit differentiation of this function because we cannot easily solve this as a y = f(x) situation. So since we have a left and right side, let's start with the right side...

d/dx (x) = dx/dx = 1 [ Power Rule Law ]

Now let's turn to the left side...

We see a function of a function so we have to use Power Rule and Chain Rule...

d/dx [ tan^3(xy^2 + y) ] = 3 * tan^2(xy^2 + y) * sec^2(xy^2 + x) * [ Derivative of inside function ]

derivative of inside function --> Using Associativity and Product Rule now...

d/dx ( xy^2 + x ) = d/dx (xy^2) + d/dx(x) = dx/dx * y^2 + x * 2 * y * dy/dx + dx/dx

d/dx( xy^2 + x ) = y^2 + 2xy*dy/dx + 1

Putting this all together gives...

d/dx [ tan^3(xy^2 + y) ] = 3 * tan^2(xy^2 + y) * sec^2(xy^2 + x) * [ y^2 + 2xy*dy/dx + 1 ]

Now let's solve for dy/dx...

3 * tan^2(xy^2 + y) * sec^2(xy^2 + x) * [ y^2 + 2xy*dy/dx + 1 ] = 1

1/( 3 * tan^2(xy^2 + y) * sec^2(xy^2 + x ) = y^2 + 2xy*dy/dx + 1

[ 1/( 3 * tan^2(xy^2 + y) * sec^2(xy^2 + x ) - 1 -y^2 ] / 2xy = dy/dx

From there you could potentially do some simplification