## Partial Derivatives.

### May 15, 2010

When we have a regular ol’ graph on the xy-axis, it’s not that hard to start taking derivatives.  You know how this goes:

1. Pick a point.
2. Pick a point close to it.
3. Make a secant line.
4. Pick a point closer and make another line.
5. Eventually, our secant lines will converge to a tangent line at our original point, if the function is sufficiently nice.

In general, then, if a derivative exists, it will be given by

$\displaystyle f'(x) = \lim_{(\delta x\rightarrow 0)} \frac{f(x + \delta x)-f(x)}{\delta x}$

But this formula is astoundingly beautiful if we have that our function is a polynomial.  Suppose that $f(x) = a_{n}x^n + \dots + a_{0}$.  Then what do we get?

$\displaystyle f'(x) = \lim_{(\delta x\rightarrow 0)}\frac{a_{n}(x + \delta x)^n + \dots + a_0 -(a_{n}x^n + \dots + a_{0})}{\delta x}$

And we can make this a lot nastier by doing a binomial expansion.  But suffice it to note that every term without any $\delta x$ will get canceled out, and every remaining term has at least one $\delta x$ in it.  Factoring this one out and canceling it from the top and the bottom, then taking the limit as $\delta x\rightarrow 0$ we have that the only things which are left are those terms which had exactly one $\delta x$ in them after we expanded the binomial.  These terms are exactly those of the form

${i \choose i-1}x^{i-1}$

for each power.  But, of course,

${i \choose i-1} = i$

This gives us exactly these terms:

$na_{n}x^{n-1} + (n-1)a_{n-1}x^{n-2} + \dots + 2a_{2}x + a_{1}$

which is a really sweet deal.  So, for polynomials, the deal is to just “pull down the power”, “multiply”, and “subtract 1 from the power.”

But something crazy happens when we have more than two variables. Specifically, this time, we’re going to be talking about what happens when our function is a surface in the xyz plane — or, in other words, without loss of generality (and up to a rotation) we’ve got a function $f(x,y) = z$.

For example, this graph is $f(x,y) = x\sin(x)$.

Now, let’s try to take a derivative!  Let’s follow those steps that we wrote down before.  Okay, picking a point isn’t that hard.  Any point will do.  Now, let’s pick a point and make a secant line.  Okay, good, now, let’s take a few of them, closer to our original point.  Okay, wait a second now.  What if we take our points coming in from some other direction?  Will we get the same thing?  Look at one of those points on the right side of the graph “going down the hill.”  If we choose our points going “up and down the hill”, our “derivative” will be something like –2.  But if we choose them going “to the left and right” of our point on the hill, the derivative will just be 0, as the secant lines are all horizontal.  We’ve got two totally different derivatives for a single point.  What do we do?

Well, we can do the next best thing.  Let’s find the derivative in the x-direction (by choosing only colinear points parallel to the x-axis), and similarly for the y- and z-direction.  We call these partial derivatives and they’re defined nearly the same way as other derivatives.  We usually write them as $\frac{\partial}{\partial x}$ for the x-direction, and similarly for the other directions.

You might think it’s really tough to find this kind of derivative given some function!  But you’d be wrong.  Dead wrong.  It’s really easy!  Here’s the algorithm:

1. Given some function $z = f(x,y)$, in order to find the partial derivative in the x-direction we let the variable y stand as a constant, and then take the derivative as usual.
2. To find the partial in the y-direction, let the variable x stand as a constant, and then take the derivative as usual.

Yeah, that’s it.  Let’s just do three examples to solidify this:

1. Let $f(x,y) = x + y$.  Then $\frac{\partial f}{\partial x}(x,y) = 1 + 0 = 1$.  Similarly $\frac{\partial f}{\partial y}(x,y) = 0 + 1 = 1$.
2. Let $f(x,y) = yx^2 + 2y$.  Then $\frac{\partial f}{\partial x}(x,y) = 2xy + 0 = 2xy$.  Similarly $\frac{\partial f}{\partial y}(x,y) = x^2 + 2$.
3. Let $f(x,y) = y\cos(x) + y$.  Then $\frac{\partial f}{\partial x}(x,y) = -y\sin(x)$.  Similarly $\frac{\partial f}{\partial y}(x,y) = \cos(x) + 1$.

Good.  Take note that the partial derivatives in either directions are still functions of x and y; that is, the derivatives in either direction will most likely depend on both the x and the y coordinate of the point we’re taking the partial derivative at!

But this is cute.  We have a way to see “approximately” what the stuff around our point looks like.  At the very least, we can see in the x and y direction.  This is like standing on a hill and only being able to look directly forwards or backwards, left or right, but never on a diagonal.

Our next post we will remedy this problem, and show how to take a derivative in a general direction (hint: there’s a way to write it as a partial in the x and y direction!).  We will also talk about how to find maximums and minimums on surfaces — when they exist, what they look like, and what the determinant has to do with them!  Exciting stuff, truly.