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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The issue is that, since the slope is 2x, it changes as you change x. At x = 1, the slope is 2; at x = 1.0000000001, the slope is 2.0000000002; and so on for any tiny changes in x.

If you wanted to calculate what the position was using the slope, you’d have to calculate it at a high precision, and to do it iteratively (use slope = 2 to find y at x = 1.00000001, recalculate the slope, then use it to find y at x = 1.00000002, and so on). You can’t easily do the math out algebraically for this; this is why calculus was invented, to deal with these situations.

A derivative is an *instantaneous* rate of change. The important concept to understand is that just because the derivative at x = 2 is 4, that doesn’t mean that this rate of change is “locked in” at 4 for all other values of x. The function y = x^2 doesn’t have a constant rate of change, but rather it increases faster and faster as x gets larger and more positive. Instead, when x is 2, the *instantaneous* rate of change, for that single point, is increase 4 y for every 1 x. But as soon as you go to a different x, even just a bit bigger than 2, the instantaneous rate of change also changes (and increases). And so on, for all values of x. By the time we get to x = 3, the instantaneous rate of change has increased from 4 to 6, which reflects the fact that x^2 increases faster and faster as x gets larger.

Here’s a real world example of the same idea: when you drive in your car and push down on the gas pedal, the car accelerates. At a particular moment during that acceleration, your *instantaneous* rate of acceleration might be an increase of 10 miles per hour, but a split moment later, your foot pushes down harder on the pedal and now your instantaneous rate of acceleration might be an increase of 20 miles per hour. It doesn’t stay the same because when you pushed down on the gas pedal, it caused a *smooth and continuous* acceleration, and at every point in time when you were speeding up, the instantaneous rate of acceleration was changing.

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.

The derivative is not 2dx, it is 2x. Your example is using the former. If we go from 2 to 3 then we need to add 2x to y and x starts at 2. So 2^2 + 2×2 = 8 which is pretty close to 9.

Now by the time we reach x=3 the derivative has gone up to six so to be more accurate we can use the value of the derivative that is between x=2 and 3 for the whole trip. The average of 4 and 6 is 5. Using 5 as the derivative we can go from x=2 to x=3 as 2^2 + 5 = 9. This time we arrived at exactly 3^2!

X^2 is X multiplied by itself? If X=4 you multiple 4×4 if X=69 you do 69×69 or if X=1 you do 1×1 similarly if you do X^3 you do X times X times X

4x4x4

Y’all downvoted need but I just tried to answer the question I thought was being asked