## Friday, November 15, 2013

### Friday Fun Day

For my calculus class, Fridays we meet in a different classroom. I like the change of scenery and it always makes me think about doing some extra fun, irregular exercise things on Fridays with the kids.

They've been working a lot on Chain Rule, but ironically only with polynomials embedded in polynomials, polynomials embedded in sine and cosine, and vice versa. They've also done an itty bitty related rates. We haven't seen the product or the quotient rule, and just the Chain Rule has kept us really busy.

Anyhow, on this Friday Fun Day, I thought of introducing the idea that reciprocal trig functions can be differentiated using Chain Rule. I told them that in reality, they won't often use those formulas, but after my hint at the Chain Rule, they were able to successfully differentiate f(x) = csc(x) and g(x) = sec(x).

Then, with only 10 minutes left in the class, I asked them to pair up and to assign one person as partner A, and the other as partner B. I asked each person to pull out their graphing calculator and to graph y=e^x. Then, they together pick an x value. Partner A needs to find f(x) and Partner B needs to find f'(x) using the calculator, and to compare answers to see what they notice.

In about 60 seconds, one kid said, "This is a trick!!"

haha. But, I wasn't prepared for their question that followed, as to WHY e^x is its own derivative. After class I did a bit of research and thought I'd post it here in case you Calculus teachers out there are wondering the same. The standard proof involves knowing how to differentiate ln(x), but as we haven't gotten there yet, I think this is a better explanation:

e is a value that comes from continuous compounding formula, namely the part (1 + 1/n)^n, limit taken as n approaches infinity.

e^x is therefore the limit as n approaches infinity of (1 + 1/n)^(nx)

By binomial expansion (which unfortunately my kids have never seen), this looks like:
1^(nx) + (nx choose 1) 1^(nx - 1)(1/n) +  (nx choose 2) 1^(nx - 2)(1/n)^2 + (nx choose 3) 1^(nx - 3)(1/n)^3 + (nx choose 4) 1^(nx - 4)(1/n)^4 + ....

= 1 + (nx)(1/n) + [(nx)(nx - 1)/2] (1/n)^2 + [(nx)(nx - 1)(nx - 2)/(2*3)] (1/n)^3 + [(nx)(nx - 1)(nx - 2)(nx - 3)]/(2*3*4)(1/n)^4 + ....

As n approaches infinity, this becomes

e^x = 1 + x + x^2/2 + x^3/(2*3) + x^4/(2*3*4) + ...

So if you differentiate each term with respect to x, you get:

d(e^x)/dx = 0 + 1 + x + x^2/2 + x^3/(2*3) + .... which is the same as the original sequence.

BAM. It's a trick. ;)

PS. I am mad that I tried to tell my husband on Friday about my e^x discovery, and he said off-handedly (before I gave him any mathematical details), "Oh yeah, I remember expanding e^x and then each term in the expansion has a derivative term that maps back to the original expansion." DAMN. He's 32 and doesn't remember most things most times; why does he still remember this?!?! That's just not right.

PPS. Yes, we have this sort of conversations. It's very geeky.

#### 1 comment:

1. It is also interesting to look at the graphs of y=a^x and its derivative for different values of a. If 1e it will be above. Therefor, the must be a value for which they are the same.