User:A Delicious Jam/Calculus AB Basics

The two main things you learn about through AP Calculus AB are derivatives and integrals. Both of these rely on the concept of a limit, and both are extensions of concepts you should already be familiar with: slope and area (respectively).

Derivatives
Just as linear functions have slopes, other functions have something similar to slope, known as a derivative. Graphically, it is represented as the slope of a line tangent to a given curve, as seen on the right. You are probably used to (or sick of) seeing the slope formula by now, where $$m = \frac{\Delta{}y}{\Delta{}x}.$$ Finding the derivative is basically taking that slope formula and making it infinitely small. Isaac Newton cleverly devised a different way to write the slope of a formula so he could apply it to curves. You can compare the usual way people would write the slope formula to Isaac Newton's way of writing it below:
 * $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{f(x + h) - f(x)}{(x + h) - (x)}$$

As it stands, this easily applies to any linear function either way. The only difference is probably one that should have been stressed in precalc, namely writing "y" as "f(x)". By doing that, you can take any x value and add any h value to a certain linear function and get the same value. You can even go a step further and just simplify it to
 * $$m = \frac{f(x + h) - f(x)}{h}$$

Since $h$ is basically the x distance (or the "run" of "rise over run") between the two points anyway.

So you may be asking yourself, "how does all this clever algebraic trickery apply to calculus?" Well, as a slight aside, the most difficult part of AP Calculus is going to be algebra. All of the concepts are intuitive enough but it's the math behind it that'll get to you. Now that we've got that out of the way, the only way to get a perfect tangent line is if the line touches the curve at exactly one point. Since a mathematical point has no length, width, or depth, that means we have to get the slope of a section of the curve that has distance zero. Okay, that may sound kinda shaky, but let's try and use this crazy limit thing and try to just plug in pure numbers from there:
 * $$m = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \frac{f(x) - f(x)}{0} = \frac{0}{0}$$

...That didn't seem to work. And how could it possibly work? Isn't it always just gonna equal 0/0? But this is exactly where all the crazy algebra comes in. Let's take the "derivative" of a linear function, say f(x) = 2x, and see if we can do some finagling to get the answer we already know is correct.
 * $$m = \lim_{h \rightarrow 0} \frac{2(x + h) - 2x}{h}$$

We can use the distributive property here and find
 * $$m = \lim_{h \rightarrow 0} \frac{2x + 2h - 2x}{h} = \lim_{h \rightarrow 0} \frac{2h}{h} = \lim_{h \rightarrow 0} 2 = 2$$

So that worked, but so far we haven't seen anything we haven't seen before. We already know that the slope of 2x is 2, and that's never gonna change. But where the calculus finally comes in is when you try to find the derivative of a non-linear function. Let's do an easy one to start, like f(x) = x2.
 * $$m = \lim_{h \rightarrow 0} \frac{(x + h)^2 - x^2}{h}$$

Let's expand that pesky (x + h)2:
 * $$m = \lim_{h \rightarrow 0} \frac{x^2 + 2xh + h^2 - x^2}{h} = \lim_{h \rightarrow 0} \frac{2xh + h^2}{h} = \lim_{h \rightarrow 0} (2x + h) = 2x$$

So now that we've stepped out of the comfortable realm of linear functions, suddenly we get a second function! And if you plug in any x value into that function, you will get the slope of f(x) = x2 at that particular x coordinate. As a quick, completely unrigorous and awful sanity check, if you were to draw a tangent line to x2 at x = 0, the tangent line would be perfectly horizontal, which would mean it would have a slope of 0, which happens to also be 2 times 0.

This function used to describe the derivative of another function is commonly expressed as f'(x), assuming the original function is f(x). If it's g(x), then you use g'(x), and hopefully you are sensing a pattern here. You tend to find people do not ever use m when finding the derivative of a function, so use f'(x) instead, or consider that, if slope is Δy/Δx, then derivative, which is just a tiny version of slope, is just dy/dx. Since there is a tendency for AP test writers to use different words to describe the same thing, know that derivative, slope of the tangent line, and instantaneous rate of change all refer to this stuff.

That's all the advice I'm willing to share to start you off on your journey, try using this formula (known as the definition of the derivative):
 * $$\lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$$

to find the derivative of other functions (I would suggest sticking to polynomials) and maybe see if you can find any patterns.

Integrals
I wasn't taught the nitty-gritty on integrals until around November or December so this section will not be as in-depth as the previous one because what I learned required a lot of background information with regards to derivatives. But the gist of it is that integration is to differentiation as multiplication is to division, in that they are opposites of each other. This idea is known as the fundamental theorem of calculus.

Rather than finding the slope of the curve, you will be asked to find the area between the curve and the x-axis (at least initially), bounded by vertical lines or even other functions. Before you learn more precise methods, you might start approximating using rectangles and triangles to approximate the area under a curve. Using more and more rectangles to approximate the area becomes more and more precise, and as the width of the rectangle approaches 0, you end up finding the definite integral, expressed as this:
 * $$\int_a^b f(x)\,dx$$

The squiggly line is the integral operation, which is basically a sum (hence why it looks kinda like an S) of infinitely narrow strips (which also means the "dx" has the same use here as it does in "dy/dx": an infinitely small change in x) with height f(x) from x coordinate a to x coordinate b. In basic terms, you're finding the area of a curvy shape bounded by f(x), two vertical lines, and the x-axis, as shown on the right.

If the formula does not have the a and b, it is an indefinite integral, less confusingly known as an antiderivative, which you can occasionally find by reversing the process you use to find the derivative once you learn various rules and shortcuts. Just know there are a lot of functions where you won't be able to do that in AP Calculus AB, and those will probably be given to you as definite integrals to do on your calculator.

Here's an example of a very basic definite integral:
 * $$\int_4^5 5\,dx$$

Is the area from x=4 to x=5 of f(x) = 5. The height of each infinitely narrow strip is 5 throughout the entire unit-long region of the graph. This essentially forms a rectangle with length 1 and height 5, so the area, and therefore the definite integral, of the region is just 5. When you start studying integrals you will tend to use rectangles and triangles to estimate the area of a region before you get into the technical details behind exact measurements.

Other stuff to expect

 * You won't be using your calculator a whole lot, because the class is algebra heavy so you won't really need it much of the time. A majority of the questions on the test are either on the No Calculator portion or they were written by college professors who hate calculators enough to sneak problems that make using a calculator take longer on the Calculator Allowed portion.
 * If you haven't gotten into the habit already, you should write the exact answer even if it looks as stupid as $$\frac{e^{17} + \pi}{x^2}.$$ If that isn't possible, round your answer to exactly three decimal places.
 * If your previous teachers haven't gotten on your ass about showing your work, they have failed as math teachers. The FRQs require you to show all setup and work, and honestly, it's just better to find out when you made a silly algebra mistake (which you will do, about a thousand times, especially on the Volumes of Revolution unit)