# What is the derivative of x^2y^2?

Mar 24, 2015

I will assume that you want $\frac{d}{\mathrm{dx}} \left({x}^{2} {y}^{2}\right)$

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

We are assume that $y$ is some function or functions of $x$. Think of it as

${x}^{2} {y}^{2} = {x}^{2} \cdot {\left(\text{some function}\right)}^{2}$

To differentiate this, you'd need the product rule and the chain rule.

d/(dx)[x^2*("some function")^2]=2x*("some function")^2+x^2*2("some function")*"the derivative of the function"

That looks kinda complicated, but we have names for the $\text{some function}$ and for its derivative. We call them $y$ and $\frac{\mathrm{dy}}{\mathrm{dx}}$.

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

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If my assumption was mistaken:

If you wanted to find the partial derivatives of the function $f \left(x , y\right) = {x}^{2} {y}^{2}$, then you will not need the product rule.

$\frac{\partial}{\partial x} \left({x}^{2} {y}^{2}\right) = 2 x {y}^{2}$
because ${y}^{2}$ is constant relative to $x$

and

$\frac{\partial}{\partial y} \left({x}^{2} {y}^{2}\right) = 2 {x}^{x} y$
because ${x}^{2}$ is constant relative to $y$