Question

Implicit differentiation in terms of x and y

Answer 1

For implicit differentiation, you are using the chain rule
$f'(x) = g'(u(x))* \frac{du}{dx}$
Except u(x)  = y, So after every "y" term is differentiated it will be multiplied by dy/dx.

17) $y^3 +4 = 3x \\ 3y^2 \frac{dy}{dx} +0 = 3$
Then you solve for dy/dx as if its a variable.
$\frac{dy}{dx} = \frac{3}{3y^2} = \frac{1}{y^2}$

18)  Here lets review product rule:
$(fg)' = f'g + fg'$
Take derivative of each term
$2x^2 = -3y^3 +3x^2y^3 \\ 4x = -9y^2 \frac{dy}{dx} + (6x)(y^3) + (3x^2)(3y^2 \frac{dy}{dx}) \\ 4x = -9y^2 \frac{dy}{dx} +6xy^3 +9x^2y^2 \frac{dy}{dx}$
Solve for dy/dx using factoring:
$4x - 6xy^3 = -9y^2 \frac{dy}{dx}+9x^2 y^2 \frac{dy}{dx} \\ 4x - 6xy^3 = \frac{dy}{dx}(-9y^2 +9x^2 y^2) \\ \frac{dy}{dx} = \frac{4x - 6xy^3}{-9y^2 +9x^2 y^2}$