# How do you differentiate 2y^2+2x^2=5?

Apr 18, 2015

I assume that you are looking for the derivative of $y$ with respect to $x$ -- that is, for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

$2 {y}^{2} + 2 {x}^{2} = 5$

$\frac{d}{\mathrm{dx}} \left(2 {y}^{2} + 2 {x}^{2}\right) = \frac{d}{\mathrm{dx}} \left(5\right)$

Now, remember that $y$ is some function of $x$ that we have left implicit rather than making explicit. (We have not 'solved for y', but we could have.)

So, $\frac{d}{\mathrm{dx}} \left({y}^{2}\right) = 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$ (We're using the Chain Rule here.)

$\frac{d}{\mathrm{dx}} \left(2 {y}^{2} + 2 {x}^{2}\right) = \frac{d}{\mathrm{dx}} \left(5\right)$

$4 y \frac{\mathrm{dy}}{\mathrm{dx}} + 4 x = 0$

And $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y}$

(Notice that we could have written $4 y \frac{\mathrm{dy}}{\mathrm{dx}} + 4 x \frac{\mathrm{dx}}{\mathrm{dx}} = 0$, but, since $\frac{\mathrm{dx}}{\mathrm{dx}} = 1$, there's not much point.)