### 8.1 This is why partial derivatives are so easy...

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- You take a picture of some atmospheric phenomenon—a cloud, wind-blown dust, precipitation, haze, winds blowing different directions—anything that strikes you as interesting.
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In your first calculus class you learned about derivatives. Suppose we have a function *f* that is a function of *x*, which we can write as *f*(*x*). How does f change with a very small change in x? This change is quantified by the derivative of f with respect to x.

What about a new function that depends on two variables, *h*(*x,y*)? This function could, for example, give the height *h* of mountainous terrain for each horizontal point (*x,y*). So what is the derivative of *h* with respect to *x*? One way we determine this derivative is to fix the value of *y *=* y _{1}*, which is the same as assuming that

*y*is a constant and then taking the ordinary derivative of

*h*with respect to

*x*. Suppose we take a slice through the mountain in the

*x*-direction at a fixed value of

*y*=

*y*that is

_{1},*, y*kept constant. How does

*h*vary with a small change in

*x*on this slice? This change is quantified as

This is called the **partial derivative** of *h* with respect to *x*. It’s pretty easy to determine because we do not need to worry about how *y* might depend on *x*.

Similarly, we can find the partial derivative of h with respect to y. In this case, we assume that *x* is constant and then take the ordinary derivative of *h* with respect to *y*. This change is quantified as

Are the partial derivatives of *h* with respect to *x* and to *y* the same? Try different functions for *h(x,y)* to see if you can find one for which they are equal for all values of *x* and *y*. (hint: Try *h(x,y)* = a constant.) Can you find any others?

#### Check Your Understanding

Let $h={\left(x-3\right)}^{2}\mathrm{cos}\left(y\right)$ . What is the partial derivative of *h* with respect to *x*?

**Click for answer.**

$$\frac{\partial h}{\partial x}=\frac{\partial \left({\left(x-3\right)}^{2}\mathrm{cos}(y)\right)}{\partial x}=2\left(x-3\right)\mathrm{cos}(y)$$

We can also find the partial derivative of *h* with respect to *y*. Can you do this?

**Click for answer.**

$$\frac{\partial h}{\partial y}=\frac{\partial \left({\left(x-3\right)}^{2}\mathrm{cos}(y)\right)}{\partial y}=-{\left(x-3\right)}^{2}\mathrm{sin}(y)$$

So you can see that the $\partial h/\partial x$ may be different for each value of *y* and $\partial h/\partial y$ may be different for each value of *x*. Thus, even if you are not entirely familiar with partial derivatives and their notation, you can see that they are no different from ordinary derivatives but you take the derivative with respect to just one variable at a time.

#### Need more practice?

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