# How do you use implicit differentiation to find (dy)/(dx) given 4x^2=2y^3+4y?

Nov 27, 2016

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

#### Explanation:

We differentiate as usual BUT remembering to include the term $\frac{\mathrm{dy}}{\mathrm{dx}}$ when we differentiate any $y$; so we get:
$8 x = 6 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 4 \frac{\mathrm{dy}}{\mathrm{dx}}$
Collect $\frac{\mathrm{dy}}{\mathrm{dx}}$:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8 x}{6 {y}^{2} + 4} = \frac{8 x}{2 \left(3 {y}^{2} + 2\right)} = \frac{4 x}{3 {y}^{2} + 2}$

Nov 27, 2016

#### Explanation:

For the x term, you can just use the power rule:

$\frac{d \left(4 {x}^{2}\right)}{\mathrm{dx}} = 8 x$

For the y terms, you use the power rule but then multiply by $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{d \left(2 {y}^{3} + 4 y\right)}{\mathrm{dx}} = \left(6 {y}^{2} + 4\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

NOTE: You are really using the chain rule along with the power rule.

Put the equation back together:

$8 x = \left(6 {y}^{2} + 4\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8 x}{6 {y}^{2} + 4}$

Remove $\frac{2}{2}$:

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

Done.