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You should review those five sections, review your notes, and look over the examples that we've done in class, your quizzes, your homework, etc. Things on the exam should look familiar.
I've told you flat out some of the things that you're going to see on this, if you've been paying attention.
I don't do study guides, because students have weaponized them: if I ask something that's not explicitly on the study guide, students will complain that I've done something unfair. So why would I tie my hands?
Let's do a few more examples.
To prove it we use one of my favorite tricks: the addition of a special form of 0.
where $n$ is a positive integer.
We can prove this via dominoes -- i.e., mathematical induction -- or via the binomial theorem)
The derivative of the monomial $x^n$ is $nx^{n-1}$, and
The derivative of the monomial $c x^n$ is $nc x^{n-1}$ (by constant multiple).
A polynomial is just a sum of these. So we apply the sum rule, and the power rule, and the constant multiple rule to the flight of the eraser, to get
$s'(t)=(at^2+bt+c)'=2at+b$
and
$s''(t)=(2at+b)'=2a$
The general rule for a polynomial $P(x)$, where is \[ P(x)=\sum_{k=0}^n c_{k}x^k \] is \[ P'(x)=\left(\sum_{k=0}^n c_{k}x^k\right)'=\sum_{k=0}^n (c_{k}x^k)'=\sum_{k=1}^n c_{k}kx^{k-1} \] Notice that the 0-term disappears, because it's a constant, and a constant's derivative is 0 (so that term "disappears").
$\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$
where $n$ is any non-zero real number (although we won't prove that).
Arc Length: $L = \theta r$
Again, this is an example of a linear relationship, but it's called "bi-linear", because it's linear in two different ways: $L$ is proportional to each of $\theta$ and $r$. So
Now you might wonder if there should be, or could be, an extra constant in there, but you can see that the formula works out for the circumference formula for a circle: \[ \theta=2\pi, \textrm{ so } L=2 \pi r \]
Yep! That's the formula for the circumference of a circle.
We can use this diagram (and the arc length formula) to prove a really important limit: \[ \lim\limits_{\theta \to 0}{\frac{\sin(\theta)}{\theta}}=1 \]
(See page 142, which I'll walk you through now. The proof uses some elementary trigonometry, plus the squeeze theorem, which is one of the really cool things about this proof!)
We will derive the derivative of the sine function from the limit definition of the derivative (although we can see the derivative graphically above).
There are exactly three important trig identities one needs to know (all the others can be derived from these three):
So let's see how to derive the derivative of the sine from the MIDIC -- the Most Important Definition In Calculus -- the limit definition of the derivative -- using the second of these identities, plus one more important limit: \[ \lim\limits_{\theta \to 0}{\frac{\cos(\theta)-1}{\theta}}=0 \]
Then the derivative of the cosine can be derived by simply shifting the sine function, and using the second trig identity above.