# How do you find the derivative of xy^2?

Sep 14, 2015

If you want to differentiate this expression as part of an implicit differentiation problem, here is how:

#### Explanation:

Assuming that we want to find the derivative with respect to $x$ of $x {y}^{2}$ (assumong that $y$ is a function of $x$:

First use the product rule:

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

Now for $\frac{d}{\mathrm{dx}} \left({y}^{2}\right)$ we'll need the power and chain rules.

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

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

If you want to differentiate the expression with respect to $t$ then the derivatives above are all $\frac{d}{\mathrm{dt}}$ and $\frac{\mathrm{dx}}{\mathrm{dt}}$ may not be $1$.

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

If you want the partial derivatives of the function $f \left(x , y\right) = x {y}^{2}$ that has been answered in another answer.