# What is the derivative of sqrt(xy)?

##### 2 Answers
May 15, 2015

It depends on which partial derivative you're interested in. Let's develop some possibilites of derivation.

For our purposes, let's just rewrite it as $z = {\left(x y\right)}^{\frac{1}{2}} = {x}^{\frac{1}{2}} \cdot {y}^{\frac{1}{2}}$

Let's find the first partial derivative for $x$:

$\frac{\partial z}{\partial x} = \left(\frac{1}{2}\right) \cdot {x}^{- \frac{1}{2}} \cdot {y}^{\frac{1}{2}} = \left(\frac{1}{2}\right) {\left(\frac{y}{x}\right)}^{\frac{1}{2}}$

Now, in the same sense, for $y$.

$\frac{\partial z}{\partial y} = \left(\frac{1}{2}\right) \cdot {y}^{- \frac{1}{2}} \cdot {x}^{\frac{1}{2}} = \left(\frac{1}{2}\right) {\left(\frac{x}{y}\right)}^{\frac{1}{2}}$

Edit:

Expanding our discussion, as this is a really common production function in Microeconomics (but also used, with some variations, as some Macroeconomics' growth models - even some Nobel prize winners started with a Cobb-Douglas), we can interprete these results.

Considering, for example, a production function where $x$ stands for the input of capital and $y$ stands for the input of labour, then the first derivative of each will give us their respective marginal product, that is, how much an unitary increase from each input will result in terms of increase in our output (production).

It is important to underline that the exponents may vary from case to case - or from an economic sector to another. For example, it is possible to have a production function such as $z = {x}^{0.1} \cdot {y}^{0.2}$ as well as in other cases, something like $z = {x}^{3} \cdot {y}^{4}$.

The cool part of it is that when you sum the exponents you get the nature of the company/sector returns: generalizing the function as $z = {x}^{\alpha} \cdot {y}^{\beta}$ when $\alpha + \beta > 1$ you have economies of scale; when $\alpha + \beta < 1$, you have diseconomies of scale and when $\alpha + \beta = 1$, you have constant returns to scale.

May 15, 2015

If this is one part of a bigger implicit differentiation problem, here's the derivative of this one term with respect to $x$:

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

Method:

I've used:

$\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{1}{2 \sqrt{u}} \frac{\mathrm{du}}{\mathrm{dx}}$ (With $u = x y$)

And the product rule to find:

$\frac{d}{\mathrm{dx}} \left(x y\right) = \frac{d}{\mathrm{dx}} \left(x\right) \cdot y + x \cdot \frac{d}{\mathrm{dx}} \left(y\right) = 1 y + x \frac{\mathrm{dy}}{\mathrm{dx}}$