The solutions to Practice Exam #2 have been posted.

Study hard. Be ready to go on time Friday.

I will post on Wednesday's lecture after I say a few more things on Thursday!

In Section 3.7, I probably should've assigned problems in the range from 19 to 29. Some of those would be good to examine before Friday's test. (That's a hint ...)

Monday in 121 we introduced a kind of expansion of the derivative-taking we've been doing. This differentiation works to give us the instantaneous rate of change in the case that our curve is not the graph of a function, but of an equation containing x's and y's. Since y is not given as an explicit formula involving x, we say that y is implicitly understood to depend on x (but we can't say exactly how). Finding the derivative in this case is called implicit differentiation.

The technique itself is not difficult: We take the derivative across the entire relational equation. Every time we touch an x-term, we take the derivative as usual. Every time we touch a y-term, we have to tack on a due to the Chain Rule. Getting the hang of this seemed within most folks' grasp -- we often just have to decide when and where to invoke the Product, Chain, or Quotient Rules. The difficult part can often be the algebra involved in solving the resulting equation for .

We had an example on an equation of a circle to show that the new method and old method agreed; we also did an example on the Folium of Descartes, where there was no way to find the needed tangent line slope via the old methods.

Tomorrow we work on applications from implicit derivatives.

Exam #2 happens this Friday, October 10th. It covers Sections 3.1 through 3.7 -- that is everything we've had on derivatives thus far.

Here are some practice problems for the exam. They are due on Friday at the exam, but I will post the solutions on Thursday.

The solutions to Quiz #5 are now posted in the Test and Quiz Solutions page.

Friday in Calc I, we started our discussion of higher derivatives by treating them as a mechanical exercise, something we did just because we could, and we introduced notation.

We provided motivation by appealing to an application. If is the position for an object at time t, then one of the motivations for finding the derivative was that it represented the velocity . The acceleration is then defined to be the instantaneous rate of change of the velocity. That is,

.

Using Leibniz notation can often help us keep track of units. We can write

.

So, for instance, if s is in meters, and t is in seconds, the notation for v clues us in that v is in m/sec, and we can think of a as being in (m/sec)/sec or the equivalent m/sec².

We did a fairly quick example involving position, velocity, and acceleration. We then moved on to generic higher derivatives. We can notate the nth derivative as



We saw that for s, a position function, the third derivative is called the jerk, the instantaneous rate of change of acceleration. We also saw that for f, an nth degree polynomial, if we take more than n derivatives, we get a value of 0. (That doesn't happen in general.)

Usually to compute, say, the 256th derivative of a function, we have to compute the 255 derivatives that came before. Sometimes, we can get lucky and get an explicit formula for the nth derivative in general. We saw this with .

Monday we talk about how to find rates of change along graphs that aren't graphs of functions. See you then.