# How do you implicitly differentiate x+xy-2x^3 = 2?

May 24, 2015

Let us define $f \left(x , y\right) = x + x y - 2 {x}^{3} = 2$.

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$, and so every time you differentiate ($\frac{\mathrm{df}}{\mathrm{dy}}$) a part of the function that has $y$, you have to multiply by $\frac{\mathrm{dy}}{\mathrm{dx}}$, the derivative of $y$ with respect to $x$, for the overall derivative ($\frac{\mathrm{df}}{\mathrm{dx}}$) to still be with respect to $x$ (even though you're differentiating $y$).

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{d \left(x\right)}{\mathrm{dx}} + \left[x \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + y \cdot \frac{d \left(x\right)}{\mathrm{dx}}\right] - \frac{d \left(2 {x}^{3}\right)}{\mathrm{dx}} = \cancel{\frac{d \left(0\right)}{\mathrm{dx}}}$
(has product rule)

$1 + \left[x \frac{\mathrm{dy}}{\mathrm{dx}} + y\right] - 6 {x}^{2} = 0$

Simple algebra from here:
$x \frac{\mathrm{dy}}{\mathrm{dx}} = 6 {x}^{2} - 1 - y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 {x}^{2} - y - 1}{x}$