A few administrative notes before we review the day’s material: I will not be holding office hours this Wednesday. And there are no classes next Monday, when my usual set of office hours are. But I’ve decided to do a sort of experiment: I don’t plan on reviewing for the exam specifically next week, but a large portion of the class has said that they would come to office hours on Monday if I were to have them. So I’m going to hold them to that – I’ll be in Kassar House 105 (the MRC room) from 7-8:30 (or so, later perhaps if there are a lot of questions), and this will dually function as my office hours and a sort of review session.

But this comes with a few strings attached: firstly, I’ll be willing to answer any question, but I’m not going to prepare a review; secondly, if there is poor turnout, then this won’t happen again. Alrighty!

The rest is after the fold –

The topic of the day was differentiation! The three questions of the day were –

- Differentiate the following functions:
- $latex e^x$
- $latex e^{e^x}$
- $latex e^{e^{e^x}}$
- $latex \sin x$
- $latex \sin (\sin x)$
- $latex \sin (\sin (\sin x))$

- A particle moves along a line with its position described by the function $latex s(t) = a_0t^2 + a_1t + a_2$. If we know that it’s acceleration is always $latex 20$ m/s/s, that its velocity at $latex t = 1$ is $latex -10$ m/s, and its position at $latex t = 2$ is $latex 20$ m. What are $latex a_0, a_1, a_2$?
- Given that $latex u(x) = (x^2 + x + 2$, what are the following:
- $latex \frac{d}{dx} (u(x))^2$
- $latex \frac{d}{dx} (u(x))^n$
- $latex \frac{d}{dx} (5 + x^3)^{-3}$
- $latex \frac{d}{dx} ((u(x))^n)^m$

#### Question 1

This is all about the chain rule. Please note that this is a big deal, so if you have any trouble at all with the chain rule, seek extra help. The derivative of $latex e^x$ is $latex e^x$. To compute the derivative of $latex e^{e^x}$, we might think of $latex u(x) = e^x$, so that we have $latex e^u$. The derivative of $latex e^u$ will be $latex e^u u’$, which gives us $latex e^{e^x}e^x$. Let’s look at the other way of understanding the chain rule to compute the derivative of $latex e^{e^{e^x}}$. The “outer function” is $latex e^{(\cdot)}$. It’s derivative is just itself. The first “inner function” is $latex e^{e^x}$. We have just computed its derivative above (it’s $latex e^{e^x} e^x$). So we multiply them together to get $latex e^{e^{e^x}}e^{e^x}e^x$.

Similarly, the derivative of $latex \sin x$ is $\cos x$. The derivative of $\sin \sin x$ requires the chain rule. On the one hand, the outer function is $latex \sin$, and the derivative of $latex \sin$ is $\cos$. So we know we will have a $latex \cos (\sin x)$ in the answer. The inner function is also $latex \sin x$, so we need to multiply by its derivative. The final answer will be $latex \cos (\sin x )\cos x$. To compute the derivative of $latex \sin \sin \sin x$, we again use the chain rule. I will again use helper functions, to illustrate their use. We might call $latex u(x) = \sin \sin x$, so that we are computing the derivative of $latex \sin (u)$. Then we get $latex \cos u u’$. We happen to have computed $latex u’$ just a moment ago, so the final answer is $latex \cos \sin \sin x \cos \sin x \sin x$.

#### Question 2

The key idea of this question is to remember that the function $latex s(t)$ gives position at time $latex t$. So its derivative gives a result in terms of position per time, the velocity. And the derivative of velocity will give a result in terms of position per time per time, or acceleration. So the velocity of our particle is $latex 2a_0t + a_1$, and the acceleration is $latex 2a_0$. Since we know that the acceleration is always $latex 20$, we know that $latex 2a_0 = 20$ so that $latex a_0 = 10$. The velocity at $latex t = 1$ is $latex -10$, so we know that $latex 2(10)(1) + a_1 = -10$, so that $latex a_1 = -30$. Finally, our position at time $latex t = 2$ is $latex 20$, so that $latex 4(10) + 2(-30) + a_2 = 20$, so that $latex a_2 = 40$. I used different numbers between the two classes, so don’t pay too much attention if the exact details are different between one class and the other.

#### Question 3

This is more about the chain-rule! This is sort of an explicit example of helper functions. We first want to compute the derivative of $latex u(x)^2$. By the chain rule, this will be $latex 2u(x)u'(x)$. What is $latex u'(x)$?. It’s $latex 2x + 1$. So the derivative of $latex u(x)^2$ is $latex 2(x^2 + x + 2)(2x + 1)$. This is a single case of the slightly more general $latex u(x)^n$. Here, the power rule tells us that the derivative will be $latex nu(x)^{n-1}u'(x)$, which is $latex n (x^2 + x + 2)^{n-1}(2x + 1)$.

The idea behind the third question is to see if we can work out the same sort of idea, but without starting with a helper function. (It’s perfectly fine to always use helper functions to use the chain rule – that’s not a problem at all). The derivative of $latex (5 + x^3)^{-3}$ will be $latex -3(5 + x^3)^{-4}(3x^2)$. If we want to see the use of helper functions, call $latex v(x) = 5 + x^3$, so that we are computing the derivative of $latex v^{-3}$. The derivative will be $latex -3v^{-4}v’$, which is exactly what we have above.

I look forward to seeing some of you on Monday, and happy studying!

I’m writing on behalf of a emailed question that I feel might be appropriate for more people to see:

This comes in two parts.

Let’s find the points of continuity of $latex csc(2x)$. We know that $latex csc(2x) = dfrac{1}{sin 2x}$, so we are wondering about when $latex sin (2x) = 0$

To do this, we first need to remember when $latex sin x = 0$. The zeroes of $latex sin x$ are $latex x = 0, pm pi, pm 2 pi, …$ or $latex x = n pi$ for any integer $latex n$ (this includes $latex sin pi = 0$, which you mentioned in your email). But we now have $latex 2x$ instead of $latex $, so we might ask ourselves when $latex 2x = n pi$ for any integer $latex n$. This happens when $latex x = dfrac{n pi}{2}$.

Let’s check ourselves – $latex sin (2 frac{n pi}{2}) = sin (n pi) = 0$. So this seems about right.

You ask: why do we only care about when $latex tan x$ is continuous when we consider $latex dfrac{x tan x}{x^2 + 1}$? To answer this, we need to remember something from the book:

If $latex f(x)$ and $latex g(x)$ are both continuous at a point $latex x = c$, then their product $latex f(x)g(x)$ is also continuous at the point $latex x = c$.

For our problem, we might realize that $latex dfrac{x}{x^2 + 1}$ is continuous everywhere, but $latex tan x$ is only continuous whenever $latex x neq frac{pi}{2} + pi n$ for any integer $latex n$. This means that their product $latex dfrac{x tan x}{x^2 + 1}$ is continuous at least when both $latex tan x$ and $latex dfrac{x}{x^2 + 1}$ are continuous. So our function is continuous at least when $latex x neq frac{pi}{2} + pi n$.

But what about when $latex x = frac{pi}{2} + pi n$? Let’s look at one in particular: say $latex x = pi/2$. $latex tan pi/2$ is not defined (the left limit is $latex + infty$, the right limit is $latex – infty$, but $latex tan x$ has a vertical asymptote there). In fact, $tan x$ isn’t defined at any of these points $latex x = frac{pi}{2} + pi n$, so these points are not in the domain of our function. So the function can’t be continuous there.

So the discontinuities of $latex dfrac{x tan x}{x^2 + 1}$ are exactly the places where $latex tan x$ is discontinuous.

During my office hours, I mentioned that this not because of anything particularly special about $latex dfrac{x}{x^2 + 1}$, but simply because $latex dfrac{x}{x^2 + 1}$ is continuous. In a sense, multiplying by a continuous function never reduces the points of continuity.