Basic Differentiation Rules

Return to Course Outline

Previous Lecture

Anything You Want to Talk About?

(No.)

Misc

Rejuvenation Day

Next Wednesday is the semester’s first “rejuvenation day,” i.e., a day without classes, intended as a break from academics. We won’t have a class meeting that day, and the cohort schedule will shift one day to accommodate — I’ll be careful to reflect that in announcements next week. I also won’t have a problem set to do during next week — though if you grade early in the week you’ll probably still want to meet next Monday or Tuesday to show me problem set 3, and I may hand out problem set 4 Thursday or Friday, to work on and grade during the following week. You can also use next week to catch up on meetings if you need to, e.g., problem sets 1 or 2, redos, etc. But note that I’m planning to not hold meetings on Wednesday.

SI

The next session is Sunday, 6:00 - 7:30. Please go, these sessions can be a big help, particularly if you want more in-person opportunity to learn material from this course.

Problem Set 3

Don’t forget that problem set 3 is now available. You should be working on it this week and grading it late this week or early next.

Differentiation Rules

From “The Basic Rules,” “The Power Rule,” and “The Sum, Difference, and Constant Multiple Rules” in section 3.3 of the textbook, and this discussion.

Using the Rules

Find formulas for f′(x) if...

f(x) = 3x + 2

On the surface, this just requires applying what many people think of as an extension of the power rule (the rule that says the derivative of xⁿ is nx^n-1) to sums or differences of powers with coefficients.

That extended rule is really the application of a number of simpler rules presented in the book. Specifically...

Documenting the steps in taking a derivative in this much detail is unusual. More often you might identify a number of rules you can apply more or less simultaneously in order to find a derivative. For example...

Applying the Rules

In physics, speed or velocity is the rate of change of position, intuitively how fast you’re moving from place to place. If you throw a ball straight up from ground level with an initial speed of 20 feet per second, its height t seconds later is given by h(t) = 20t - 16t² feet above ground. How fast is it moving as a function of time?

Using the idea that speed is the derivative of position, and that height is basically a measure of position, we just need to find the derivative of h(t):

In economics, marginal cost is the rate of change in the cost of producing a quantity of some product. Intuitively, you can think of it as the cost of producing one additional unit of the product after already producing a very large number. For example, if it costs C(q) = 50q + 1000 dollars to produce q widgets, what’s the marginal cost of one widget?

Using the idea that marginal cost is the derivative of cost, find C′(q):

If you think of the cost function as reflecting $1000 of set-up costs (buying widget-machines, etc.) plus $50 of materials per widget, this marginal cost says that once you’ve made many widgets, the set-up costs are amortized away and the only cost of another widget is the materials it directly uses.

Proving the Power Rule

This gives an interesting insight into what exactly the power rule as we know it now says and when we can use it, but the actual proof is beyond what this course would normally do. So you won’t be tested on the following, but might find it interesting, especially if you like math and knowing where it comes from.

Assume we don’t already know the power rule, we’re trying to derive it, or prove that it’s correct. So the only relevant thing we have for finding the derivative of xⁿ is the limit definition:

Our general goal in finding this limit (as well as many others) is to get rid of the thing that makes the expression inside the limit undefined when h equals the value it’s approaching. In this case that “thing” is the division by h. So we’d like to find some way to cancel it out, or otherwise get rid of it. Expanding the numerator is the natural, and indeed maybe the only, way towards this.

To expand the numerator, we particularly need to expand the (x+h)ⁿ part. There is a general rule for expanding such things, based on the so-called “binomial coefficients”:

Using this expansion produces...

Now, motivated by wanting to cancel away the h in the denominator, and the fact that lots of the terms in the numerator now have a factor of h in them, see if there’s some way to get rid of all the terms in the numerator that don’t contain h. And in fact, there is: there are only 2 such terms, xⁿ and -xⁿ, which conveniently cancel each other out:

Now we can factor h out of all the remaining terms in the numerator and cancel it with the h in the denominator:

Now we’ve achieved the immediate goal, getting rid of the h in the denominator. Having done that, see if we can just substitute h = 0 into the remaining expression and evaluate it. We can, making all but the first term in the sum vanish (because they all involve multiplication by h):

Finally, use the equation for binomial coefficients to find the one we have left:

This proof has a couple of morals:

1. It’s an example of how everything in mathematics (in this case a differentiation rule) has a reason (here, following from the definition of derivative).
2. It relied on expanding a power of a binomial, using a formula that requires the exponent to be a positive integer. Because it relied on that, we do not now have a power rule that applies to negative exponents, fractional ones, etc. This is a good example of where “fine print” in mathematical theorems can come from.

Next

Miscellaneous topics related to derivatives, including

• Derivatives of derivatives (higher-order derivatives)
• Running differentiation rules in reverse (antiderivatives)
• Derivatives in Mathematica
• Derivatives and graphs of functions.

Please read “Graphing a Derivative” and “Higher-Order Derivatives” in section 3.2.

Please also participate in this discussion about some of the above topics.

Next Lecture