# How do you differentiate x^2y + xy^2 = 6?

Apr 18, 2015

I assume that you are trying to find $\frac{\mathrm{dy}}{\mathrm{dx}}$. We think of $y$ as a function (or some functions) of $x$, but rather than solving for $y$ to make the function explicit, we leave the function implicit.

We will differentiate using the chain rule. So $\frac{d}{\mathrm{dx}} \left(y\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$ and $\frac{d}{\mathrm{dx}} \left({y}^{2}\right) = 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$ .

For this problem we will also need the product rule, because ${x}^{2} y$ is 'really'
${x}^{2} \left(\text{some function of x}\right)$ like ${x}^{2} \left(g \left(x\right)\right)$ and similarly for the other term $x {y}^{2}$which is 'really' $x \cdot {\left(s o m e f u n c t i o n\right)}^{2}$

${x}^{2} y + x {y}^{2} = 6$. Because these are equal, the derivatives must be equal:

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

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

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

$2 x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + {y}^{2} + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

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

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

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