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

## Thanks for your help!

Aug 5, 2016

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

#### Explanation:

When we implicitly differentiate a function $f \left(x , y\right) = 0$, whenever we differentiate w.r.t. we use chain rule and when we differentiate w.r.t. $y$ and then multiply it by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

As such as we have $3 y + {y}^{4} / {x}^{2} = 2$

$3 \times 1 \times \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{4 {y}^{3} \times \frac{\mathrm{dy}}{\mathrm{dx}} - {y}^{4} \times 2 x}{x} ^ 4 = 0$ or

$3 \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x {y}^{4}}{x} ^ 4 = 0$ or

$3 \frac{\mathrm{dy}}{\mathrm{dx}} + 4 {y}^{3} / {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {y}^{4} / {x}^{3} = 0$ or

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[3 + \frac{4 {y}^{3}}{x} ^ 4\right] = \frac{2 {y}^{4}}{x} ^ 3$ or

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

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