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Note, however, that this topic will not be covered on the exam.
Important Note: you will be expected to show your work for the exam! So, while you might use a calculator, say, to compute a derivative, you will have to be able to show me how you got it using our rules.
Note one more thing, that we didn't mention last time: while it is true that \[ \sin'(x) = \cos(x) \] one can also see that \[ \sin'(x) = \sin\left(x+\frac{\pi}{2}\right) \]
This is a trig identity ($\cos(x)=\sin\left(x+\frac{\pi}{2}\right)$ -- so we don't even really need two functions -- one is just a shift of the other); but it says something really interesting: differentiation is like a shift operator for sine.
Sine looks down the road by $\frac{\pi}{2}$ for the slope of the tangent line at a point. Can you see that? And isn't that weird?:)
Note that this figure shows that the derivative is consistent with the graph of the function:
Theorem (the chain rule): If $F(x) = f(g(x))$, and both $f$ and $g$ are differentiable at $x$, then \[ F^\prime(x) = f^\prime(g(x))g^\prime(x) \]
I personally think about the chain rule this way:
"f prime of stuff times stuff prime.",
\[ F^\prime(x) = f^\prime(stuff)stuff^\prime \]
You can see that the rule is fairly simple, once you've identified the composition -- that is, once you've torn apart $F$ to find $f$ and $g$.
You might take a look at this summary from my pre-calc class to review compositions.
Before we do that, however, I'd like to motivate (rather than prove) the chain rule, using the limit definition of the derivative. Everything comes from that!
Basically, however, it relies on the tangent line: we want to use the fact that \[ g(x+h) \approx g(x) + hg'(x) \] Why does that make sense? Because it comes straight out of the limit definition, where we throw away the limit. That's why we have to write "$\approx"$:
\[ g'(x) \approx \frac{g(x+h)-g(x)}{h} \]
If time were measured in years from January, would
be a good model? What would be a good choice for the parameter $A$?