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

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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*). What is 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 take the ordinary derivative of

*h*with respect to

*x*. In a sense, we are taking a slice through the mountain in the

*x*-direction at a fixed value of

*y*=

*y*. Thus,

_{1}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*.

#### 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 for just of one variable at a time.

#### Need more practice?

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