Day 14, Mat128: Calculus A

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Announcements:
- If you need to Zoom to class, here's the link.
- I hope that you had a good and relaxing spring break!
- Today we continue to use our tools to compute some derivatives of more trig functions.
- You've got a new homework assignment:
  1. Read section 2.5 (the chain rule); do the preview and submit prior to class next time.
  2. Reminder that you have an IMath problem set on the go....
- Recent whiteboard work:
  - Today's
Last time:
- we built the following rules, creating even more useful new tools out of our old (but good) tool, the limit definition of the derivative (plus the new rules we just built!): \[ F'(x)=\lim_{h \rightarrow 0}\frac{F(x+h)-F(x)}{h} \] Again, for the following proofs, assume that functions $f$ and $g$ are differentiable at $x$.
  - Theorem (quotient rule -- not simply the quotient of the derivatives!):
    Let $F(x)=\frac{f(x)}{g(x)}$. \[ \frac{d}{dx}[F(x)]=\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{\frac{d}{dx}f(x)g(x) - f(x)\frac{d}{dx}g(x)} {[g(x)]^2} = \frac{f'(x)g(x) - f(x)g'(x)} {[g(x)]^2} \] I got a rhyme from a colleague, to help us remember this formula: ``Lo dee hi minus hi dee lo, over the denominator square they go!''.
    But we'll obtain this theorem by using tools we've already built; in particular, the product rule, and a trick to get the derivative of the multiplicative inverse of a function.
    A theorem that you prove on the way to proving some theorem is called a "lemma".
    - Lemma (derivative of the multiplicative inverse):
      Let $F(x)=\frac{1}{g(x)}$. Then \[ \frac{d}{dx}[F(x)] = \frac{-g'(x)}{g(x)^2} \]
  - Theorem (exponential rule):
    Let $F(x)=e^x$. \[ \frac{d}{dx}[F(x)] = F(x) \hspace{1in} (= F'(x) = F''(x) = F'''(x) = ....) \]
    where $e \approx 2.71828$ (it's irrational, like $\pi$).
    This is certainly one of the most amazing rules. It says that $e^x$ is an exponential function which is its own derivative: whose values are its slopes, as well (and all its higher derivatives as well -- its concavity, its jerk, ...).
    Here's an animation which motivates the discussion...
  - Theorem (trigonometric derivatives):
    Let $F(x)=\sin(x)$. \[ \frac{d}{dx}[F(x)]=\cos(x) \]
    We'll need a couple of limits to establish this one, as well as one trig identity: \[ \sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a) \]
  - Corollary:
    Let $F(x)=\cos(x)$. \[ \frac{d}{dx}[F(x)]=-\sin(x) \]
    This is a consequence of the derivative of sine, by symmetry and periodicity.
    Further consequences: \[ \frac{d^2}{dx^2}[\sin(x)]=-\sin(x) \] \[ \frac{d^3}{dx^3}[\sin(x)]=-\cos(x) \] \[ \frac{d^4}{dx^4}[\sin(x)]=\sin(x) \]
    The sine (and cosine) functions are their own fourth derivatives.
- Today's worksheet just asks us to produce derivatives and graphs of other : Section 2.4 worksheet.
Links:
- The source of the animation of the exponential functions. Thanks to conroy....

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Updated on 03/16/2022 04:38:33