At x=2, y is x^2 , which is 4. at x=3, y is 9. If y is changing at a rate of 2x for every change in x, wouldn’t that make y at x=3 6, because you moved 1 along the x so you move 2 along y? Or how does this work? I’m having trouble understanding differentiation 🙁
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Let’s start by working with only the integers. If we have x^2, this creates a function like so:
f(0) = 0
f(1) = 1
f(2) = 4
f(3) = 9
Now, what is the slope at f(2)? Well, we know it must be between (4-1)/1 = 3 and (9 – 4)/1 = 5 because those are the slopes from f(1) to f(2) and f(2) to f(3).
So let’s look a little closer:
f(1.5) = 2.25
f(2.5) = 6.25
So we now know that our slope must be between (4 – 2.25)/0.5 = 3.5 and (6.25 – 4) / 0.5 = 4.5.
Just these two iterations should give you a clue that if we continue this process, taking points closer and closer to 2, we’ll end up somewhere around 4 (which is 2x).
Now, let’s look at it another way. What’s the slope of x^1? Well, this is a straight diagonal line and the slope of a straight diagonal line is constant. So we know the derivative of x^1 must be some function with x^0 in it (if it has any other power of x, it won’t be constant).
Back to x^2. If you examine the results above, we can approximate the slope just from the integer values. We know the slope from f(0) to f(1) is 1, the slope from f(1) to f(2) is 3, the slope from f(2) to f(3) is 5. These differences are both the upper and lower bound for our actual slope (the upper bound is just one later in the sequence) and it should be obvious that this is an additive sequence.
In other words, our derivative of x^2 must have be a function with x^1 in it – because that’s how we generate a sequence where each subsequent value is 2 more than the previous.
We can make the same observation for any positive power of x^n, although it’s obviously a bit more difficult to see.
For a slightly more comprehensive view, consider the following equation of the slope between two points on the x^2 curve an arbitrarily small distance k apart:
slope = ((x + k)^2 – x^2) / k
This is simply the standard rise/run you learned back in algebra.
Simplify a bit:
(x^2 + 2x^1k + k^2 – x^2) / k
2x^1 + k
Since k is arbitrarily small, we’ll just ignore it and get:
2x^1
Now, the mathematicians in the audience are wincing a bit because you can’t really do what I did above without using limits, but it should give an idea of how the math resolves.
Note: You can continue this process for any x^n. Consider x^3:
((x+k)^3 – x^3) / k = (x^3 + 3kx^2 + 3k^2x + k^3 – x^3) / k
The x^3 cancel just like the x^2 cancelled above, all the terms with k^2 or greater fade to insignificance and all you’re left with is that term with a power one less than the power you started with.
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