(Please read parts 1 & 2 first)

From what we learned in parts 1 & 2, the derivative is the instantaneous slope of a line at a point. So, how do we calculate it? To do this, we need to first redefine the slope formula.

To most, the formula for a slope is (y2 - y1) / (x2 - x1). Now, for us, we know what the notation for functions is (so y1 is really f(x1) and y2 is f(x2) ), so we can rewrite the formula as ( f(x2) - f(x1) ) / (x2 - x1). Now, let us define "h" to be the difference between x1 and x2, meaning h = x2 - x1. Then, using substitution, the formula becomes ( (f(x2) - f(x1) ) / h. And, since h = x2 - x1, that means x2 = x1 + h (adding x1 to both sides). Then, we can substitute again to make our final version: slope = ( f(x1 + h) - f(x1) ) / h.

Now, since h is the difference between the x-values, to get the instantaneous slope (the derivative), we want h to be zero (from part 2). We cannot divide by zero though, so we say: derivative (D) = "the limit as h approaches 0 of ( f(x1 + h) - f(x1) ) / h". This formula is the algebraic definition of the derivative.

Okay, so the derivative is a fancy rewrite of the slope formula with a limit attached... so what? Well, let us return to our example from part 2: f(x) = x^2, where we tried to find the slope at x = 1. I will be leaving out "the limit as h approaches zero"
in my equations in order to keep them shorter but realize the limit is still there:

1. By our definition of the derivative, D = ( f( 1 + h) - f(1) ) / h.

2. Now, replace the function notation with the actual function (f(x) = x^2) and we get D = ( (1 + h)^2 - (1)^2) / h.

3. Factoring gives us D = ( (1 + 2h + h^2) - (1) ) / h.

4. We can cancel out the ones to get D = (2h + h^2) / h.

5. Factoring out an h gives us D = (h*(2 + h))/h.

6. Now, we can cancel out the h as long as h is not zero. Since we are using the limit of h as it approaches zero, technically h itself is not zero, so we can do the cancellation to get that the derivative, D = 2 + h.

7. Since we now no longer have division, we can approximate the limit to be zero itself, which gives us 2 + 0 = 2.

8. Thus, the derivative of f(x) = x^2 at x = 1 is 2.

I know what we just did is probably a bit confusing. It is much harder to write this than to explain in person. Here is the basics of what we did:

1. We rewrote the slope formula to use function notation and used h = x2 - x1.

2. We applied the formula to the point in question, substituting the value for x1 in the derivative equation and then using the function to determine what the numerator was.

3. We used algebra to simplify the numerator until we were able to factor out an h (which is always possible to do).

4. Using that the limit meant we were looking at h being very close to zero but not actually zero, we canceled the h from the numerator and denominator.

5. Realizing that now we no longer have a division problem, we justified that since h was very close to zero that we could replace h with zero and solve the equation.

As they are probably the hardest to understand, I will try to explain the difference between steps 4 & 5. The difference between the x-values of our two points (what we called "h") could not be zero since that would imply that the two points were actually the same, so step 4 was legitimate. At the same time, we want h to be as close to zero as possible, and since step 5 no longer had division, we could use substitution, and the closest thing to zero is indeed zero, so we used zero as an estimate of h, so step 5 was legit.

Now that you hopefully have an idea of the derivative, I will mention that there is also the derivative function. Instead of finding the instant slope at a point, the derivative function gives the instant slope at any point. Let us again take a look at f(x) = x^2.

1. Start with the derivative formula, but this time use x itself as the "x1": D = ( f(x + h) - f(x) ) / h

2. Now apply the function: D = ( (x + h)^2 - (x)^2) / h

3. Factor: D = ( (x^2 + 2xh + h^2) - x^2) / h

4. Cancel the two x^2: D = (2xh + h^2) / h

5. Factor out the h: D = (h*(2x + h))/h

6. Since we are using the limit (even though I have not been writing it), we can cancel the h: D = 2x + h

7. I want to rewrite this properly now with limit: The derivative is "the limit as h approaches zero of 2x + h"

8. Approximate h to be zero: the derivative of f(x) = x^2 is f ' (x) = 2x + 0 = 2x

When writing the derivative function, we use the ' to denote that this is the derivative. The derivative function is actually much more powerful than the derivative at a point. This is because it gives us the instant slope at ANY point. So, if we wanted to find the derivative at x = 1, we have f ' (1) = 2(1) = 2, which is precisely what we got before.

I hope that this blog trilogy has helped to clarify what the derivative is. I may revise or rewrite this series later if I can find better ways to use words to describe what I am doing.