Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 152: 10

$y' = \frac{12x^2 + 1}{4y^3 - 2}$

$\frac{d}{dx}(y^4)-\frac{d}{dx}(2y) = \frac{d}{dx}(4x^3) + \frac{d}{dx}(x)$ $4y^3y' - 2y' = 4*3x^2 + 1$ Now we have to solve for $y'$: $y'(4y^3 - 2) =12x^2 + 1$ $y' = \frac{12x^2 + 1}{4y^3 - 2}$