Day 22, Mat128: Calculus A

Last Time

Next Time

Announcements:
- Your quiz this week will be over inverse functions.
Last time::
- Let's start with 3.2
- We were working on inverse functions, including the important examples of $\ln(x)$, arcsin; and at the end of the hour, arctan and arccos.
- Let's look over the quiz over the chain rule.
  - Key
  - Some things to work on:
    1. Simplification (#1)
    2. Parentheses (#2)
    3. Simplification (#4) -- and paying attention to domains
  - Some solutions:
    1. Emily
    2. Joseph
    3. Gabriel
    4. Caroline
Today:
- We begin with consideration of section 3.2 ("Using derivatives to describe families of functions"), where we will reflect a little on compositions while thinking about parameters.
- Then let's return to those last two inverse function: arctan and arccos. Just as before, but once again we have to restrict the domain: cosine and tangent are not invertible. We have a choice, but it seems like the best place to think of tangent as invertible is on the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$:
- Now how about its derivative? For $f(x)=\tan(x)$, $f'(x)=\sec^2(x)$ -- still easy! But this time it will be a little more unpleasant to compute $f'(f^{-1}(x))$. There's a trick you know, however: we can rewrite $\sec^2(x)$ in terms of tangent, using our favorite trig identity: \[ \sin(x)^2+\cos(x)^2=1, \] Dividing by $\cos^2(x)$ we get \[ \tan^2(x)+1=\sec^2(x), \] Hence \[ \frac{d}{dx}\left(arctan(x)\right) = \frac{1}{\sec^2(arctan(x))} = \frac{1}{1+\tan^2(arctan(x))} = \frac{1}{1+x^2} \] Thus \[ \frac{d}{dx}\left(\arctan(x)\right) = \frac{1}{1+x^2} \]
- Now let's do it for arccos: we have a choice, but it seems like the best place to think of cosine as invertible is on the interval $[0,\pi]$:
- Now how about its derivative? You might recognize that it looks a lot like the arcsin function: in fact, we can see that it is the reflection of arcsin about the y-axis, followed by a shift up by $\pi/2$: hence \[ \arccos(x)=\arcsin(-x)+\frac{\pi}{2}. \]
  That means we can get its derivative by the chain rule: \[ \frac{d}{dx}\left(\arccos(x)\right) = \frac{-1}{\sqrt{1-x^2}} \]
- Here's a Section 2.8: using derivatives to Evaluate limits worksheet for us to work on (a bit of "review" of the inverse trig functions, too -- let's all take a look at those together, to start).
  The first page is about inverse trig functions, so let's take a look at that "review" for a moment (we haven't covered some of it!).
- And finally, for today's new trick, we'll consider something called "L'Hôpital's rule" (Section 2.8: Using Derivatives to Evaluate Limits)
  
  Suppose $h(x)=\frac{f(x)}{g(x)}$, where $f$ and $g$ are differentiable and $g'(x) \ne 0$ near $a$ (except possibly at $a$). Suppose that \[ \lim_{x \to a}f(x)=0 \;\;{\text{and}}\;\; \lim_{x \to a}g(x)=0 \] Then \[ \lim_{x \to a}h(x)= \lim_{x \to a}\frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)} \] if the derivatives on the right side exists, and $g'(a) \ne 0$.
  Now let's see how to derive this (or to make sense of it, at any rate): since $f(a)=g(a)=0$, we can write \[ \lim_{x \to a}h(x) = \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f(x)-f(a)}{g(x)-g(a)} = \lim_{x \to a}\frac{\frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}} = \frac{\lim_{x \to a}\frac{f(x)-f(a)}{x-a}}{\lim_{x \to a}\frac{g(x)-g(a)}{x-a}} = \frac{f'(a)}{g'(a)} \]
- Examples:
  - Activity 2.8.2. Evaluate each of the following limits. If you use L'Hôpital's rule, indicate where it was used, and be certain its hypotheses are met before you apply it:
    1. $\displaystyle \lim_{x \to 0} \frac{\ln(1 + x)}{x}$
    2. $\displaystyle \lim_{x \to \pi} \frac{\cos(x)}{x}$
    3. $\displaystyle \lim_{x \to 1} \frac{2 \ln(x)}{1-e^{x-1}}$
    4. $\displaystyle \lim_{x \to 0} \frac{\sin(x) - x}{\cos(2x)-1}$
- Links:
  - Chain Rule summary
  - Implicit differentiation summary
  - A Keeling Analysis in Mathematica
  - Case Study: Early Blooming of the cherry trees in the nation's capital
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