Topics within Calculus

Okay, so now you're a whiz at limits, but you still don't feel like you're doing Calculus!

Yeah, I know - you want to start writing $\frac{dy}{dx}$ and those weird S-shaped signs, that look like $\int$. I know I did!

In that case, my friend, we have some good news - we're not so far from all that. Just hang in there.

On with the lesson.

Suppose we have a function $y=3x-5$.

What's the slope?

**It's 3, right?**

Good. Just checking.

Okay, next question. What's the slope of the line $$y=x^2?$$

What do you mean, you don't know?

Oh... you mean it keeps changing? Right. When x increases at a constant speed, the speed at which y increases increases.

Go back, and read that to yourself a few more times, until it makes sense.

In the graph above, the blue line is *tangent* to the parabola at the green point, which is called the *point of tangency.*

Try to move the green "tangent" point around and see the change in the slope of the line. Once you get a feel for what's going on, unhide the black line.

If we were to zoom in an infinite amount to the point of tangency, we'd have difficulty telling the line and parabola apart, because they get so close together.

In fact, you can try it yourself by clicking in the lower right of the graph and zooming in!

Because the line and the *portion* of the parabola are almost the same, we say that the slope of the parabola at the point of tangency is equal to the slope of the tangent.

Now, the only task we have is to develop a general equation for the slope of the parabola at any point.

**Note: **As you may have seen elsewhere, $\Delta$ usually stands for "change in". In Calculus, $\delta$ (lowercase $\Delta$) usually means "a very small change in". That's $\delta$ a definition you already know.

In the graph below, we have a zoomed-in version of $y=f(x)$, an arbitrary function. We've chosen two values of $x$:

$a$ and $a+\delta x$.

Their respective $y$ values on the graph will be $f(a)$ and $f(a+\delta x)$.

Okay, we're trying a new input style here. To type $\delta$ ("change in"), type "delta", and the symbol should pop up on its own.

Make sure to write your answers in terms of $f(x)$, $a$, and $\delta x$.

The general formula for slope is .

The slope of the line going through the green and blue points can be expressed as .

The second one was a bit tricky. The answer is $$\frac{f(a+\delta x)-f(a)}{\delta x}.$$

The above expression describes the change in the $y$-value, or $f(x)$, and divides it by $\delta x$, or the change in $x$.

If we can make $\delta x$ infinitesimally small, then this can help us find the slope at any value of $(x,f(x))$.

**I'm going to go ahead and swap out $x$ for $a$ in the above expression, just to emphasize that this works for any value of $x$.**

$$\frac{f(x+\delta x)-f(x)}{\delta x} = \frac{\delta y}{\delta x}.$$

Ah, yes. Where were we? We were going to compute the above when $\delta x$ is infinitesimally small. How do we do that?

Remember that *infinitesimally small* means that the number is approaching $0$, but it's not exactly 0. So we have to evaluate the above when $\delta x$ is approaching $0$.

We just learned exactly how to do that using limits!

We can set up the limit like this:

$$\lim_{\delta x \to 0}\frac{f(x+\delta x)-f(x)}{\delta x}.$$

Believe it or not, that's it!

When we plug in a definition of $f(x)$, such as $f(x)=x^2$ to the limit, we can evaluate the limit to get another function, which we call $f'(x)$.

**Quick check:** What is the relationship between $f'(x)$ and $f(x)$?

The definition of $f'(x)$ goes like this:

The value of $f'(a)$, for any $a$ within the domain of $f(x)$, evaluates to the instantaneous slope of the function $f(x)$ when $x=a$.

Basically, if $f(x)$ is your velocity, then $f'(x)$ is your acceleration.

Now, let's try an example. Let's try to find the slope of $f(x)=x^2$.

Step 1: Set up the limit. $$\lim_{\delta x \to 0}\frac{(x+\delta x)^2-x^2}{\delta x}.$$

Step 2: Solve the limit! $$\lim_{\delta x \to 0}\frac{x^2 + 2x\delta x + (\delta x)^2-x^2}{\delta x}$$

$$=\lim_{\delta x \to 0}\frac{2x\delta x + (\delta x)^2}{\delta x}$$

$$=\lim_{\delta x \to 0}2x + \delta x$$

And, now that $\delta x$ is out of the denominator, we can substitute $\delta x =0$:

$$\frac{\delta y}{\delta x}=2x$$