Day 25, Mat128: Calculus A

Last time:

Next time:

Today:

Announcements
- Your quiz Friday will cover section 2.2. There is an new assignment related to section 2.4.
- Tomorrow I will post the solutions to the worksheet, and to the 2.2 homework.
- I'll be zooming this meeting, and recording it, for a student who has a legitimate issue requiring it.
Last time:
- The exams were handed back, and those who wish may resubmit (with original attached) for 1/3 of the points back. These are due Friday.
- We went over problem 1 of the exam, thinking about the choice of tangent lines that would produce a good estimate for the second derivative.
- We went over observations and comments on the exam.
- Then we wrapped up our rules with a big one: the derivative of sine (and so discovered the derivative of cosine by symmetry/periodicity): \[ \sin'(x)=\cos(x) \] \[ \cos'(x)=-\sin(x) \]
  Notice that the derivative of sine (an odd function) is cosine (an even function); and the derivative of cosine (even) is odd.
  We deduced the derivative of sine using the limit definition of the derivative, one trigonometric identity, and some understanding of the behavior of a couple of limits:
- We also noted that \[ \frac{d^2}{dx^2}[\sin(x)]=-\sin(x) \] \[ \frac{d^3}{dx^3}[\sin(x)]=-\cos(x) \] \[ \frac{d^4}{dx^4}[\sin(x)]=\sin(x) \] The first is interesting as relating to oscillating phenomena, and the third is interesting, saying that the sine (and cosine) functions are equal to their own fourth derivatives. This is actually important in the stability of columns.
Today we move on to Section 2.4: Derivatives of other trigonometric functions. But, honestly, this is all but trivial now that we have so many cool rules -- so many power tools!
- Derivatives of other trig functions are now easily obtained using the quotient rule:
  - Derivative of tangent: \[ \tan'(x) = \left(\frac{\sin(x)}{\cos(x)}\right)' = \frac{1}{\cos(x)^2} = \sec(x)^2 \]
    We can derive this result in a slightly different fashion, which simultaneously provides the derivative of the secant function: use the product rule, rather than the quotient rule, and use a result that we discovered by "trickery" when deriving the quotient rule: \[ \frac{d}{dx}\left[\frac{1}{g(x)}\right] = \frac{-g'(x)} {g(x)^2} \] Let's compute the derivative using the product rule: \[ \tan'(x) = \frac{d}{dx}\left(\sin(x)\frac{1}{\cos(x)}\right) \]
    We'll end up with the identity \[ \tan'(x) = 1 + \tan(x)^2 = \sec(x)^2 \] The same trick can be used in the case of cotangent, to get a related identity.
  - Derivative of cotangent (this is also a shift -- the identify is \(\cot(x)=-\tan(x+\pi/2)\)): \[ \cot'(x) = \frac{-1}{\sin(x)^2} = -\csc(x)^2 \]
- Derivatives of secant and cosecant are similarly trivial to obtain, but here they are: \[ \frac{d}{dx}[\sec(x)] = \sec(x)\tan(x), \ \text{and} \ \frac{d}{dx}[\csc(x)] = -\csc(x)\cot(x)\text{.} \]
Let's take a look at the worksheet for this first part of chapter 2.
Then I've got a worksheet with a focus on section 2.4 for you.

Website maintained by Andy Long. Comments appreciated.

Footnotes

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