# How do you implicitly differentiate 3y + y^4/x^2 = 2?

Jul 7, 2016

I found: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {y}^{4}}{3 {x}^{3} + 4 x {y}^{3}}$

#### Explanation:

ou need to remember to derive also $y$ as a function of $x$, so, for example:
${y}^{2}$ derived will give you: $2 y \frac{\mathrm{dy}}{\mathrm{dx}}$
$3 \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{4 {y}^{3} {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x {y}^{4}}{x} ^ 4 = 0$
$3 \frac{\mathrm{dy}}{\mathrm{dx}} + 4 {y}^{3} / {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {y}^{4} / {x}^{3} = 0$
collect $\frac{\mathrm{dy}}{\mathrm{dx}}$ and rearrange:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {y}^{4} / {x}^{3}}{3 + 4 {y}^{3} / {x}^{2}} = \frac{2 {y}^{4}}{3 {x}^{3} + 4 x {y}^{3}}$