Day 22, Mat128: Calculus A

Last time:

Next time:

Today:

Announcements
- The exam Friday will cover sections 1.4-2.1, 2.3.
Last time:
- We reviewed the quiz, focusing on a few things:
  - Using a centered difference, the slope of a secant line, to estimate the derivative at a point $x=a$. \[ f'(a) \approx \frac{f(a+h)-f(a-h)}{2h} \]
  - I urged you to please forget about slope-intercept form! Use point-slope form: it shows everything going on locally at the point $(a,f(a))$: \[ y=L(x)=f(a)+f'(a)(x-a) \]
- We reviewed the derivative of an exponential function, noting that there is a special one -- the exponential function with base $e \approx 2.71828$ -- such that \[ F'(x)=\left(e^x\right)' = e^x = F(x) \] Of course all constant multiples of the function $F$ also satisfy this curious relationship: \[ \alpha F'(x)=\left(\alpha e^x\right)' =\alpha e^x = \alpha F(x) \] but basically there is one special function with this amazing property -- that it is its own derivative: $e^x$.
- We finished off our basic rules with the quotient rule,
- Theorem (quotient rule -- not simply the quotient of the derivatives!):
  Let $F(x)=\frac{f(x)}{g(x)}$. \[ \frac{d}{dx}[F(x)]=\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{\frac{d}{dx}f(x)g(x) - f(x)\frac{d}{dx}g(x)} {[g(x)]^2} = \frac{f'(x)g(x) - f(x)g'(x)} {[g(x)]^2} \] which we derived in two different ways:
  - using the limit definition, and
  - as a corollary of the product rule.
- Today we begin with a review of anything that you want to ask about; I'll summarize the topics, howeever:
  - 1.4: The derivative function
    - We continue using the limit definition of the derivative, only now we generate a function (rather than the derivative value at a single point).
    - We related the shape of the graph of the derivative function to that of the function itself (I asked you to become blackbelts at this exercise.
  - 1.5: Interpreting, estimating, and using the derivative
    - Don't forget the units in real problems.
    - Centered difference: This quantity measures the slope of the secant line to $y = f(x)$ through the points $(a-h, f(a-h))$ and $(a+h, f(a+h))$. The central difference generates a good approximation of the derivative of $f$ at $a$: $f'(a)$ -- which itself represents the slope of the tangent line at $x=a$.
  - 1.6: The second derivative
    - derivative of the derivative
    - acceleration
    - concavity
    - matching function, derivative, and second derivative
  - 1.8: The tangent line approximation
    - The local linearization is just that linear function that has the tangent line as its graph: \[ y-f(a) = f'(a)(x-a) \] then \[ y = L(x) = f(a) + f'(a)(x-a) \]
    - Using point-slope allows me to see at a glance the point about which the localization is occuring.
  - 2.1: Elementary derivative rules
    - the sum rule
    - the product rule
    - the constant multiple rule
    - the difference rule
    - the power rule
  - 2.3: the product and quotient rules
    - the quotient rule.
  - If we've got some time, we'll check out the worksheet, working with polynomials and powers (negative, and other than simple integers): Sections 2.1 and 2.3 worksheet.
    One thing to be aware of is that "surds" (those horrible square and cube root signs) really represent non-integer powers (and so the power rule can be used).
  - So here's the big news: we've got all the power we need to differentiate (easily) any rational function -- a ratio of polynomials -- and that's a huge and important group of functions.
    (By the way, the polynomials are merely a subset of the rational functions: they're the rational functions with constant polynomials for denominators.) \[ f(x)=4x^3-7x^2+5x-1=\frac{4x^3-7x^2+5x-1}{1} \]
    See? So what's the derivative of \[ F(x)=\frac{x^2-3x-4}{x-2} \]
    Derivatives of rational functions are easy now! (No more limit definition derivations for you -- at least most of the time). But we still have a few more rules to prove....
  - Next up: rules for derivatives of the trigonometric functions -- but these won't be on your exam.
    - Theorem (trigonometric derivatives):
      Let $F(x)=\sin(x)$.
      \[ \frac{d}{dx}[F(x)]=\cos(x) \]
      We needed 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.

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Footnotes

Derivative of $f$: \[ 12 x^2-14 x+5 \]
Derivative of $F$: \[ \frac{x^2-4 x+10}{(x-2)^2} \]