# How do you do implicit differentiation of y^3+x^2y+x^2-3y^2?

May 21, 2015

Each $y$ is some (unknown) function of $x$, so

to differentiate ${y}^{3}$ and $- 3 {y}^{2}$ we will use the power rule and the chain rule (a combination sometimes called 'the generalized power rule')

To differentiate ${x}^{2} y$ we'll need the product rule.
(I use the order: $\left(f g\right) ' = f ' g + f g '$ and I'll put the product in brackets $\left[. .\right]$)

$\frac{d}{\mathrm{dx}} \left({y}^{3} + {x}^{2} y + {x}^{2} - 3 {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left({y}^{3}\right) + \left[\frac{d}{\mathrm{dx}} \left({x}^{2} y\right)\right] + \frac{d}{\mathrm{dx}} \left({x}^{2}\right) - \frac{d}{\mathrm{dx}} \left(3 {y}^{2}\right)$

$\textcolor{w h i t e}{\text{sssssss}}$ $= 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + \left[2 x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}\right] + 2 x - 6 y \frac{\mathrm{dy}}{\mathrm{dx}}$

$\textcolor{w h i t e}{\text{sssssss}}$ $= \left(2 x y + 2 x\right) + \left(3 {y}^{2} + {x}^{2} - 6 y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

That's all we can do. If we had an equation, we could solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.