Course Overview
This course introduces the fundamental ideas behind differential calculus, including topics such as limits, differentiation and the various rules that come with limits and differentiation.
As you progress through this course, you will find links to related topics in other courses (such as algebra). This interconnected structure is designed to help you quickly revisit prerequisite ideas without breaking your workflow.
Limits
We must begin with a discussion of what a limit is, why we care about it, and examples of when it is useful.
If you want the fast version, below is a short video introducing limits.
Introduction to derivatives
Coming Soon
If you want the fast version, below is a short video introducing indeterminate products.
The Power Rule
Description of the power rule!
More to come!
The Chain Rule
Description of the chain rule!
More to come!
Derivatives of sinusoidal functions
Description of the the derivative for sinusoidal functions
More to come!
Indeterminant Forms
In this section we will be rehashing limits. If you have forgotten about limits, I strongly suggest that you click here to remind yourself of how to perform them. It should be no surprise, from how this course has progressed thus far, that we will find ourselves interested with the behavior of functions as they approach certain values. However, we may run into an issue in certain places, for example, suppose we are interested in the following rational function, $$F(x) = \frac{x^2-1}{x-1},$$ and specifically we want to know the value of \(F(x)\) as \(x \rightarrow 0\), thus we should write this as $$\lim \limits_{x \rightarrow 1} F(x) = \lim \limits_{x \rightarrow 1} \frac{x^2-1}{x-1},$$ of course the problem is, that as we approach 1 from either the left or the right we end up with \(F(x) \rightarrow \frac{0}{0}\). The problem that we are running into here is that \(\frac{0}{0}\) is undefined. This is one example of what is called an indeterminant form, \(\frac{\pm \infty}{\pm \infty}\) is another example, and there are a number of other indeterminant forms as we will see through the examples below. I want to take a moment to note how appropriate the name is, these forms are indeterminate because as we will see in the following sections, we can't (necessarily) determine beforehand what the answer will be, for instance \(\frac{\infty}{\infty}\) could be 1 or it could be \(\infty\) or it could be any real number you can choose depending on the functions involved!
If you want the fast version, below is a short video introducing indeterminant forms.
So how do we handle this problem of indeterminate forms? Well, we can handle it algebraically in some cases, the example given above is an ideal candidate for this. Notice that we can factor the numerator since it is the difference of two perfect squares, which will allow us to make a cancellation (we must be very cautious with these types of cancellation because for instance in this case it is only valid if \(x\neq1\), but we get "away" with that here because we are taking limits, not setting it equal to 1), $$\frac{x^2-1}{x-1} = \frac{(x+1)(x-1)}{(x-1)} = x+1,$$ which is true, as noted above, for \(x \neq 1\). This is perfectly fine from a limit standpoint, we are only asking the behavior as \(x\) approaches 1. So if we take the limit here, $$\lim \limits_{x \rightarrow 1} x+1 = 2.$$
This is nice, but as you probably guessed from my comment above "in some cases" that this method is not overly general. Indeed, in many cases using this type of trick is impossible. So instead, we will ask for a more general method, which I will discuss below called l'Hospital's rule.
l'Hospital's rule
I will start by stating l'Hospital's rule and then we will begin discussing applications. Suppose we have two differentiable functions, \(f\) and \(g\), and furthermore that \(g \neq 0\) and \(f,g\) are both differentiable on some open interval \(I\) around \(a\) (though it may not contain \(a\) itself). Additionally suppose either, \(\lim \limits_{x \rightarrow a}f(x)=0\) and \(\lim \limits_{x \rightarrow a}g(x)=0\) or \(\lim \limits_{x \rightarrow a}f(x)=\pm \infty\) and \(\lim \limits_{x \rightarrow a}g(x)=\pm \infty\) (notice the ratio of these functions will be in common types of indeterminate forms), then $$\lim\limits_{x\rightarrow a}\frac{f(x)}{g(x)} = \lim\limits_{x\rightarrow a}\frac{f'(x)}{g'(x)},$$ so long as the limit of the right hand side exists. In other words, the limit of ratio of the functions is the same as the limit of the ratio of each functions derivative. We can continue using our example from the last section and verify that we get the same answer. Recall that click here from the power rule, if we let \(f(x) = x^2-1\), then $$f'(x) = 2x.$$ Similarly, using the power rule, since \(g(x) = x-1\) then \(g'(x) = 1\). Thus according to l'Hospital's rule, we have $$\lim \limits_{x \rightarrow 1} \frac{x^2-1}{x-1} = \lim \limits_{x \rightarrow 1} \frac{2x}{1} = 2,$$ which is identically what we found above! So now we should work on some examples which we can't simply manipulate algebraically, which shows the real power of l'Hospital's rule.
If you want the fast version, below is a short video introducing l'Hospital's rule.
Suppose we have the function \(K(x) = \frac{\mbox{cos}(x)}{x-\frac{\pi}{2}}\) and we want to find the limit of this function as \(x \rightarrow \frac{\pi}{2}\). Notice we will run into the \(\frac{0}{0}\) indeterminate form. Thus, recalling derivatives of sinusoidal functions and the power rule, if we let \(f(x) = \mbox{cos}(x), f'(x) = -\mbox{sin}(x)\) and \(g(x) = x-\frac{\pi}{2}, g'(x) = 1\). Now taking the limit, $$\lim\limits_{x \rightarrow \pi/2}\frac{\mbox{cos}(x)}{x-\frac{\pi}{2}} =\lim\limits_{x \rightarrow \pi/2} \frac{-\mbox{sin}(x)}{1} = -1.$$ Notice that there was no easy algebraic manipulation to get to this result, but straightforward application of l'Hospital's rule worked beautifully!
OK, but I mentioned there were other indeterminate forms, specifically I already mentioned \(\frac{\pm \infty}{\pm \infty}\), so we will take a look at some examples there. Suppose for example, that \(Q(x) = \frac{-3x^3-5x^2}{2x^3-1}\) and we wish to understand the behavior of this function as \(x \rightarrow \infty\) (this, by the way, is often a question asked in physics, what happens as I get "very far away" from whatever pheonomena I am observing?) Notice the problem we have, if I let \(f(x) = -3x^3-5x^2\) then as \(x \rightarrow \infty, f(x) \rightarrow -\infty\) and similarly, if I let \(g(x) = 2x^3-1\) as \(x \rightarrow \infty, g(x) \rightarrow \infty\). So I have the indeterminate form \(\frac{-\infty}{\infty}\) to be specific. According to l'Hopital's rule, $$\lim\limits_{x \rightarrow \infty}Q(x) = \lim\limits_{x \rightarrow \infty}\frac{-9x^2-10x}{6x^2},$$ after applying the power rule to both the numerator and the denominator. But this time notice that I have a problem, I still get the indeterminate form \(\frac{-\infty}{\infty}\), but notice that I am getting somewhere, as the power has been reduced in each of \(f,g\). So, the natural thing to do here is to apply l'Hospital's rule again, $$\lim\limits_{x \rightarrow \infty}Q(x) = \lim\limits_{x \rightarrow \infty}\frac{-18x-10}{12x},$$ but again we have the same problem, but we are much closer now. So apply the rule yet again, $$\lim\limits_{x \rightarrow \infty}Q(x) = \lim\limits_{x \rightarrow \infty}\frac{-18}{12},$$ now that is NICE!. We see that the limit is \(\frac{-18}{12}= \frac{-3}{2}\). We should take a look at another example to flesh out our understanding. For example, imagine the function \(r(x) = \frac{x^3}{x^2}\) and imagine that we want to find the behavior as \(x \rightarrow \infty\). Well, we should be able to immediately identify that this is of the form \(\frac{\infty}{\infty}\), so we will apply l'Hospital's rule: $$\lim\limits_{x \rightarrow \infty}r(x) = \lim\limits_{x \rightarrow \infty}\frac{3x^2}{2x},$$ but we can see that leavies us with an indeterminate form still, so applying again, $$\lim\limits_{x \rightarrow \infty}r(x) = \lim\limits_{x \rightarrow \infty}\frac{6x}{2},$$ but this gets us \(\frac{\infty}{2}\) which is NOT an indeterminate form, this is just \(\infty\). So we have found that \(\lim\limits_{x \rightarrow \infty}r(x) = \infty\). To link this to what I said in the indeterminate forms section, the past 2 examples show why these forms are called indeterminate, it is because in both cases we had something of the form \(\frac{\pm \infty}{\pm \infty}\), and for one of them it turned out that this led to \(\frac{-3}{2}\) and in the other case it led to \(\infty\)! We will check out other indeterminate forms in the next section.
Indeterminate Products
Coming Soon
If you want the fast version, below is a short video introducing indeterminate products.
Directional Derivatives
Notice the level sets (contours) of the image of Mount Fuji. These represent the curves of constant elevevation (why not call that \(z(x,y)=c\), where \(c\) is the constant
elevation).
The partial derivative \(z_x\) represents the derivative (rate of change) along the \(x\) direction and \(z_y\) represents the derivative along the \(y\) direction.
But what if I want to know the rate of change along some other direction, such as at a \(45^o\) angle between \(x\) and \(y\)? This brings in the concept of the
directional derivative. Recall that at a single point \((x_0,y_0\) we define the partial derivative
$$z_x(x_0,y_0) = \lim \limits_{h \rightarrow} \frac{z(x_0+h,y_0)-z(x_0,y_0)}{h},$$
and
$$z_y(x_0,y_0) = \lim \limits_{h \rightarrow} \frac{z(x_0,y_0+h)-z(x_0,y_0)}{h},$$
where these represent the changes in the unit direction \(\mathbf{i}\) and \(\mathbf{j}\) respectively. If we now would like to know the rate of change along some unit vector
\(\mathbf{u} = \langle a,b \rangle \). In this case we need to recognize that we are "splitting" \(h\) along two directions, and they are being scaled by \(a\) and \(b\) along
those directions. That is, in the \(x,y\) plane, \(h^2 = (ah)^2+(bh^2)\) using Pythagorean theorem (thankfully we are working in a flat plane and this still applies!). This means
that \(x-x_0 = ah, y-y_0 = bh\). And so, since these represent the legs (with \(h\) as the hypoteneuse) we can define the directional derivative along this direction at
point (\x_0,y_0\) as,
$$D_u z(x_0,y_0) = \lim \limits_{h \rightarrow 0}\frac{z(x_0+ah,y_0+bh)-z(x_0,y_0)}{h},$$
if the limit exists, naturally.
More to come!