## Sunday, July 25, 2010

### Derivative of Sine

I recently taught my class the derivative rule for sine and cosine.  These are particularly easy rules to apply, and make for nice examples when doing the product, quotient and chain rules, so I like to introduce them early.  The trouble is that the proofs of the rules are rather complex.  We would never ask students to find the derivative of sin(x) using the limit definition during an exam.  So how much detail should be go into when proving these rules?

In Hughes-Hallett, very little proof is given (it is done in the exercises).  Instead they inspect the graph of sin(x) and use it to sketch a graph of the derivative.  This is a good exercise for the student anyway, and allows them to discover for themselves that the derivative is cos(x).  While this is definitely the way to start, it would be nice to give a slightly more rigorous explanation.

On the other hand, Stewart's calculus text gives quite a rigorous proof for the derivative of sin(x).  Starting with the limit definition, then using the sum formula for sine on sin(x+h), we arrive at $\lim_{h\to 0}\frac{\sin x \cos h + \cos x\sin h - \sin x}{h}$.  This can be regrouped to give $\sin x \lim_{h\to 0}\frac{\cos h - 1}{h} + \cos x \lim_{h\to 0} \frac{\sin h}{h}$.  Now to evaluate those two limits, Stewart takes a page and a half to give a (very nice) geometric argument, using the squeeze theorem.  Last semester, I walked my students through the argument.  It did not go well.  Leaving calculus to look at a geometry problem was just too much.

Luckily, there is another way: look carefully at those two limits.  Do they remind you of anything?  What if instead of the -1, we replaced that with cos(0)?  Or wrote sin(h) - sin(0) in the numerator of the second limit.  That's right, both are simply a derivative evaluated at 0.  The first is the derivative of cos(x) evaluated at 0.  Let's look at the graph of cosine.  What is the slope of the tangent line at x = 0?  Clearly it is 0.  What is the slope of the tangent line to sin(x) at x = 0?  Looks very much like 1 to me.  That, and we already evaluated that limit using approximation, and it also looked like 1.  We are then left with simply cos(x).  And that is not a surprise, since that is what the graph of the derivative looks like.

I know this is not rigorous, but it is convincing.  And more so, it reinforces the definition of the derivative.