# How do you implicitly differentiate  y^2= x^3 + 3x^2 ?

Nov 20, 2015

It depends on what you are using as the independent variable. For $t$, we get $2 y \frac{\mathrm{dy}}{\mathrm{dt}} = 3 {x}^{2} \frac{\mathrm{dx}}{\mathrm{dt}} + 6 x \frac{\mathrm{dx}}{\mathrm{dt}}$.

#### Explanation:

${y}^{2} = {x}^{3} + 3 {x}^{2}$

$\frac{d}{\mathrm{dt}} \left({y}^{2}\right) = \frac{d}{\mathrm{dt}} \left({x}^{3} + 3 {x}^{2}\right)$

$2 y \frac{\mathrm{dy}}{\mathrm{dt}} = 3 {x}^{2} \frac{\mathrm{dx}}{\mathrm{dt}} + 6 x \frac{\mathrm{dx}}{\mathrm{dt}}$.

Solve for the derivative you're looking for.

If the independent variable is taken to be $x$, then we get:

${y}^{2} = {x}^{3} + 3 {x}^{2}$

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

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} + 6 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} + 6 x}{2 y}$