A Few Thoughts on the Derivative

Robert, a friend and colleague of mine from the University of Tennessee, has written multiple posts on automobiles. In a most elegant prose, he describes how the car is personified and how it eventually becomes a part of who you are. It's your best mate, your friend, something that you respect and adore. Your car can even shape your personality, thoughts and ideas. It speaks to you.

This is where I draw the inspiration for this post. While cars are certainly a passion of mine, they don't generate even close to the same emotions as it does for my friend. What does, however, is mathematics. When you live and breathe the subject day in and day out, you come to appreciate not only the most complicated elements of mathematics but also the simplistic notions as well. And why not? This is the medium by which we communicate with nature, mother Earth and the cosmos. It connects us, as humans, to everything that surrounds us. And just as we develop the bond with nature by utilizing the awesome power within this universal language, we also establish a significant connection with the language itself. Mathematics begins speaking to us and before we know it, we have personified the entire field. Each mathematical element becomes the friendly neighbor we interact with on a daily basis. One that's always there to provide interesting conversation that guides and assists you in all your endeavors. And each character is respected, admired and even feared all at the same time. For the most part, however, their countless complexities tend to bring a smile to your face. So which character shall we extol upon today? To the derivative: "Sing in me, Muse, and through me tell the story."

We start with how the derivative is defined. Take a function $f(x)$ and define it's derivative as \[ \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \frac{df}{dx} = f'(x). \] If the limit exists at a point $(a,f(a))$, then we may construct the tangent line at that point using the derivative evaluated at $x=a$ as the slope. Take for example the function $f(x) = \sin x$. We find that the derivative is $f'(x) = \cos x$. We'd like to construct the tangent line at $(\pi, 0)$. Since $f'(\pi) = \cos(\pi) = -1$ this now acts as my slope for the tangent line at the aforementioned point. So we find that the equation for the line is $g(x) = -x + \pi$. Graphing $f(x)$ and $g(x)$, we see the following.

Okay, so what? We can draw the tangent line at a point on the graph of $f(x)$ just by determining it's derivative, but does this provide us a bigger picture for functions in general? Take another look at the previous graph and you'll notice something. See how the slope of the tangent line was negative ($f'(\pi)=-1$)? Notice how the graph of the function $f(x)=\sin x$ is decreasing at that point? In fact, just by examining the derivative between particular intervals, we can determine if the original function is increasing or decreasing. If the derivative is strictly positive on some interval $I=(a,b)$, then the function is increasing within that interval. Similarly, if the derivative is strictly negative on $I$, then the function is decreasing.

What about the critical values of a function (i.e. where the function exhibits a local maximum or minimum)? Remember, the way the derivative is defined tells us the slope of the tangent line at any point. Therefore, it makes sense that the evaluation of the derivative at critical values would yield zero, since this is where the function changes direction. Thus, if I set the derivative equal to zero, I should be able to locate the critical points of the function.

Let us return to our function $f(x)=\sin x$. Since $f(x)$ is $2\pi$-periodic, we will just stay on the interval $(0,2\pi)$. Now take the derivative and set equal to zero. \[ f'(x)=\cos x=0, \ \ \ \text{on} \ \ (0,2\pi) \] This gives us $x = \pi/2$ and $x=3\pi/2$. Evaluating our function $f(x)=\sin x$ at these values gives us $f(\pi/2)=1$ and $f(3\pi/2)=-1$. So, we should have our critical points at $(\pi/2,1)$ and $(3\pi/2,-1)$ on the interval $(0,2\pi)$. You may see these coordinates plotted below along our function $f(x)=\sin x$.

Now that we know how to find the critical points of a function, how can we determine whether they are local maximums or minimums? Notice that when we take a second derivative, we obtain \[ f''(x)=\lim_{h \to 0}\frac{f'(x+h)-f'(x)}{h} \] just by using the definition of the derivative. Now suppose $x$ in the above is a critical value and $f''(x) > 0$. Then we have that \[ 0 < f''(x)=\lim_{h \to 0}\frac{f'(x+h)-f'(x)}{h}=\lim_{h \to 0}\frac{f'(x+h)}{h}. \] This means that for sufficiently small $h$, we have \[ \frac{f'(x+h)}{h} > 0. \] So, if $h \to 0^-$, then $f'(x+h)<0$ and we are decreasing. If $h \to 0^+$, then $f'(x+h)>0$ and we are increasing. This tells us that for $f''(x) > 0$ at a critical $x$, that point must be a local minimum. Similar results hold for when $f''(x) < 0$ in that it tells us that the critical value $x$ is a local maximum. This is what's known as the Second Derivative Test. Gathering all the information the derivative has given us, let's go one step further and see if we can't break in on the ground floor of ordinary differential equations (ODE). Simply put, an ODE is a function $F$ of $y=y(x)$ and all of its derivatives set equal to some other function $f(x,y)$ or \[ F(y,y',y'',y''',...,y^{(n)})=f(x,y). \] With the skill level that we've developed so far, let's focus on an ODE of order one \[ \frac{dy}{dx} = f(x,y) \] on some region $I=(a,b) \times (c,d)$. The goal now, is to determine a solution of the ODE. Not knowing any particular methods of solving the above equation, let's see if we can't just determine graphical qualities of $y=y(x)$. This shouldn't be too much of a stretch for us since we've been given that the derivative is set equal to some function $f(x,y)$. By evaluating $f(x,y)$ at a coordinate within $I$, we know the slope of the tangent line for $y=y(x)$ at that coordinate. Furthermore, we can construct the tangent line at that point. Now imagine if we evaluated $f(x,y)$ at multiple coordinates within $I$ and constructed multiple tangent lines. We start to get a better picture of the behavior of our solution $y=y(x)$ without finding the explicit solution. For example, take \[ \frac{dy}{dx}=\cos (x+y) \] and let's draw multiple tangent lines in the region $I=(0,2\pi) \times (0,2\pi)$ by using the evaluation of $\cos (x+y)$ at multiple coordinates within $I$ as the slope of our tangent lines. For simplicity, we shorten the length of each tangent line and construct a unit vector in the direction of the tangent line at each coordinate. What we see is the following.

By following the vectors, we now have an idea as to the behavior of our solution. We now display a stream plot below to provide an even better understanding of our solution $y=y(x)$.

So, without even solving for the solution $y=y(x)$, we have painted a picture of the function and examined its behavior. To conclude our story, let's start by recalling all the information the derivative has given us.

1.) We can now draw the graph of the tangent line at points where $f(x)$ is defined and the derivative exists.

2.) We can determine intervals of increase and decrease.

3.) We can find critical points as well as determine whether they are local maximums or minimums.

4.) Does this allow us to paint a picture of the behavior of the function as well as graph it? Yes it does.

5.) And we have broken in on the ground floor of a formidable subject just by using the above properties of the derivative.

In other words, the derivative has given us a wealth of information about our function $f(x)$. The power that this mathematical concept unleashes to the mathematician is incredible and yet we tend to forget just how lucky we are to possess it. So next time you're analyzing different functions or simply trying to graph them, don't forget: like a good neighbor, the derivative is there.