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 dydx\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=3x5y=3x-5.

What's the slope?

It's 3, right?

Good. Just checking.

Okay, next question. What's the slope of the line


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, Δ\Delta usually stands for "change in". In Calculus, δ\delta (lowercase Δ\Delta) usually means "a very small change in". That's δ\delta a definition you already know.

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

aa and a+δxa+\delta x.

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

Make sure to write your answers in terms of f(x)f(x), aa, and δx\delta x. To type δ\delta ("change in"), type "\delta", and the symbol should pop up on its own.

What is the general formula for slope?

Solution 1

To find the slope, we take rise over run, which is

Change in yChange in x.\frac{\text{Change in } y}{\text{Change in } x}.

In math, "change in" is denoted as Δ\Delta. So, we write

ΔyΔx.\frac{\Delta y}{\Delta x}.

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

Solution 1

Substituting points into the general formula for slope, we get

f(a+δx)f(a)δx.\frac{f(a+\delta x)-f(a)}{\delta x}.

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

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

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

f(x+δx)f(x)δx=δyδx.\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 δx\delta x is infinitesimally small. How do we do that?

Remember that infinitesimally small means that the number is approaching 00, but it's not exactly 0. So we have to evaluate the above when δx\delta x is approaching 00.

We just learned exactly how to do that using limits!

We can set up the limit like this:

limδx0f(x+δx)f(x)δx.\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)f(x), such as f(x)=x2f(x)=x^2 to the limit, we can evaluate the limit to get another function, which we call f(x)f'(x).

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

Solution 1

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

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

Now, let's try an example. Let's try to find the slope of f(x)=x2f(x)=x^2.

Step 1: Set up the limit.

limδx0(x+δx)2x2δx.\lim_{\delta x \to 0}\frac{(x+\delta x)^2-x^2}{\delta x}.

Step 2: Solve the limit!

limδx0x2+2xδx+(δx)2x2δx\lim_{\delta x \to 0}\frac{x^2 + 2x\delta x + (\delta x)^2-x^2}{\delta x}
=limδx02xδx+(δx)2δx=\lim_{\delta x \to 0}\frac{2x\delta x + (\delta x)^2}{\delta x}
=limδx02x+δx=\lim_{\delta x \to 0}2x + \delta x

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

δyδx=2x\frac{\delta y}{\delta x}=2x
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