Day 30, Mat128: Calculus A

Last time:

Next time:

Today:

Announcements
- You have an assignment, over section 2.6.
- You're invited! 2024 Math/Art Pop-Up Conference We'd love to have you join us on Saturday, March 30, for
  - Fun activities
  - Free lunch! (Yes, there is!)
  - Here's the link to sign up!
  - Schedule:
    - 9:00: Hyperbolic crochet
    - 10:30: Origami
    - 11:30: Lunch
    - noon: Korean drumming
- Head's up: our third exam is Friday of next week.
- Please read section 2.7 Derivatives of Functions Given Implicitly for next time, and carry out the preview. This will be the last material to be included on the exam.
Last time:
- Examined the preview exercise graphically, to see how inverse functions are related graphically: their graphs are reflections about the line $y = x$ .
- We discussed domains and ranges of inverses -- that, in general, they're different -- and that it might be best to express the relationship between them this way: if $f$ and $g$ are inverses, then
  $f (g (x)) = x$ but $g (f (y)) = y$
  Elements of one set (values $x$ ) are transported to another set (values $y$ ), by $g$ ; then $f$ takes them back home.
  Meanwhile, elements of one set (values $y$ ) are transported to another set (values $x$ ), by $f$ ; then $g$ takes them back home.
- We talked about how it's necessary in some cases to restrict the domain of a function before one will be able to invert it:
  
  But that it's so important to be able to invert some functions we will do it, and then figure out the issues later (e.g. did we actually want the negative answer to "What's the square root of 4?").
  For example, in the quadratic formula you know that there's a $\pm$ : we need both solutions that we get from the square root!
- Then we saw how the chain rule can be used to find the derivative of an inverse function (it's a composition!): $\frac{d}{d x} (f^{- 1} (x)) = \frac{1}{f^{'} (f^{- 1} (x))}$
- We studied a most important example: the inverse function of $f (x) = e^{x}$ : we'll call it $f^{- 1} (x) = \ln (x)$ . The function $f (x) = e^{x}$ is, in fact, invertible -- no need to restrict a domain!
  
  Notice the two different sets that the functions transfer elements between:
  $f : ℜ \to (0, \infty)$ $g : (0, \infty) \to ℜ$
- Now how about its derivative? For $f (x) = e^{x}$ , $f^{'} (x) = e^{x}$ -- easy!
  So $\frac{d}{d x} (f^{- 1} (x)) = \frac{1}{f^{'} (f^{- 1} (x))}$ becomes $\frac{d}{d x} (\ln (x)) = \frac{1}{e^{\ln (x)}} = \frac{1}{x} = x^{- 1}$ Amazing! This is the "missing power" -- we didn't know how to get this particular derivative using the power rule (It would have come from the power $x^{0}$ -- but that's a constant, so it's derivative is 0. This creates a mysterious connection between powers and the exponential function, and its inverse, the logarithm.)
Today:
- Let's talk about the quiz:
  - Median: 8
  - Your key
  - My key
  - Note: $\cos (\cos (x))$ is not equal to $\cos (x^{2})$
  - A number of you have forgotten the quotient rule (and used the unfortunately easy misquotient rule). You don't have to use the quotient rule, however, as I point out on the "Your key" above.
    You do have to get it right, however!:)
- Now, for more about that derivative of $\ln (x)$ :
  
  Notice that the derivative's graph is our friend the hyperbola. Notice also that this function is odd -- it should be the derivative of an even function. We can extend the $\ln (x)$ function to the left, by considering $\ln (| x |)$ :
- By the way, this allows us to now clearly understand those derivatives of other exponential functions. Consider, for example, the function $g (x) = 2^{x}$ Because we now have the inverse function of $e^{x}$ , we can write $2 = e^{\ln (2)}$ -- more generally we need to remember that, as inverses, $x = e^{\ln (x)}$ -- at least for positive values of $x$ . And, last I checked, 2 is positive: $g (x) = 2^{x} = {(e^{\ln (2)})}^{x} = e^{\ln (2) x}$ That might look a little scary, but $\ln (2)$ is just a number -- approximately 0.69. So we can now compute the derivative of $2^{x}$ easily, thinking of it as a composition -- and so using the chain rule: $g^{'} (x) = \ln (2) e^{\ln (2) x} = \ln (2) 2^{x} = \ln (2) g (x)$ More generally, if $g (x) = a^{x}$ , $g^{'} (x) = \ln (a) e^{\ln (a) x} = \ln (a) a^{x} = \ln (a) g (x)$ And the choice of $a = e$ gives that amazing function that satisfies $g^{'} (x) = \ln (e) g (x) = 1 \cdot g (x) = g (x)$
One more important example we want to consider: the inverse function of $f (x) = \sin (x)$ : we'll call it $f^{- 1} (x) = \arcsin (x)$ .
- Just as before, but once again we have to restrict the domain: sine is not invertible. We have a choice, but it seems like the best place to think of sine as invertible is on the interval $[- \frac{π}{2}, \frac{π}{2}]$ :
- The inverse of sine is called arcsine, but note that we will sometimes denote the inverse $\arcsin (x)$ as $\sin^{- 1} (x)$ . Let $f (x) = \sin (x)$ $\sin (\arcsin (y)) = y$ and $\arcsin (\sin (x)) = x$
- We use the inverse derivative rule to find the derivative of arcsine. $\frac{d}{d x} (f^{- 1} (x)) = \frac{1}{f^{'} (f^{- 1} (x))}$
  Now how about its derivative? For $f (x) = \sin (x)$ , $f^{'} (x) = \cos (x)$ -- still easy, which in this case gives $\frac{d}{d x} (\arcsin (x)) = \frac{1}{\cos (\arcsin (x))}$
- But this time it will be a little more unpleasant to simplify that denominator ( $f^{'} (f^{- 1} (x))$ ). There's a trick you know, however: we can rewrite cosine in terms of sine, using our most important trig identity: $\sin (x)^{2} + \cos (x)^{2} = 1,$ so $\cos (x) = \pm \sqrt{1 - \sin (x)^{2}} .$ (Notice that we need both $\pm$ !) Hence $\frac{d}{d x} (f^{- 1} (x)) = \frac{1}{\pm \sqrt{1 - \sin (\arcsin (x))^{2}}}$ becomes (remember that sine and arcsine undo each other, to give $x$ !) $\frac{d}{d x} (\arcsin (x)) = \frac{1}{\sqrt{1 - x^{2}}}$ where you will notice that I chose the positive "branch". Why?
  Answer: Look at sine on this interval. If it is increasing everywhere, its inverse must be increasing everywhere: and what does that say about its derivative?
- Your turn!
  There are two other functions which deserve our attention. We call their inverses "arctan" and "arccos".
  Do exactly the same thing that we've just done in these three examples, but with two other important functions:
  1. $f (x) = \cos (x)$ -- use restricted domain $[0, π]$
  2. $f (x) = \tan (x)$ -- use restricted domain $(- \frac{π}{2}, \frac{π}{2})$
Links:
- A summary from another class: The Chain Rule
- A Keeling Analysis in Mathematica
- The latest Keeling data

Website maintained by Andy Long. Comments appreciated.
Updated on 03/27/2024 03:44:55

Footnotes

Its derivative has to be positive, too.