# 8.2: This is why partial derivatives are so easy...

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?

$\dfrac{d f(x)}{d x}$

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 = y1, 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 = y1. Thus,

$\left(\frac{d h}{d x}\right)_{y=\text { constant }} \equiv \frac{\partial h}{\partial x}$

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.

Exercise

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

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

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

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