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

Dec 8, 2016

For problems like these, we must remember that we're differentiating with respect to x.

$3 + \frac{4 {y}^{3} {x}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - 2 x \left({y}^{4}\right)}{{x}^{2}} ^ 2 = 0$

$3 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + \frac{4 {y}^{3} {x}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - 2 x {y}^{4}}{x} ^ 4 = 0$

$\frac{3 {x}^{4} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 4 {y}^{3} {x}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - 2 x {y}^{4}}{x} ^ 4 = 0$

$3 {x}^{4} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 4 {y}^{3} {x}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - 2 x {y}^{4} = 0$

$3 {x}^{4} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 4 {y}^{3} {x}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2 x {y}^{4}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(3 {x}^{4} + 4 {y}^{3} {x}^{2}\right) = 2 x {y}^{4}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x {y}^{4}}{{x}^{2} \left(3 {x}^{2} + 4 {y}^{3}\right)}$

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

Hopefully this helps!