# How do you differentiate (y+3x)^2-4x=0?

Oct 19, 2015

Assuming that we want $\frac{\mathrm{dy}}{\mathrm{dx}}$, we get:

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

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

Now solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$

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

Rewrite using algebra until you like the way it looks.