I’m embarrassed to admit I’m getting my masters degree in a math related subject and I still don’t get this!
I know how to do them, but the way you compute them is almost suspiciously simple. What’s the logic behind converting the exponent to a constant? How does that determine the slope?
In: 0
Consider the function f(x) = x^2. You probably know the derivative of it is f'(x)=2x. You’re asking how this comes about? Someone else mentioned the difference quotient formula already, which is often expressed as:
the limit as h goes to 0 of (f(x+h)-f(x))/h.
Now using f(x)=x^2 and working out the algebra you can quickly get that:
the difference quotient becomes the limit as h goes to 0 of 2x+h, which is just 2x.
However, you might be wondering how the difference quotient even arises in the first place, which is really where the intuition of derivatives is held. Here’s one way to think about.
Consider again f(x)=x^2. You might want to know how far up the function goes vs how far over to the right it goes. With a linear function y=mx +b this is always the slope m, and you can calculate it by doing the change in y over the change in x between any two points.
For f(x)=x^2 the change in y over the change in x will not always be the same, it depends on the 2 points you pick. This change in y over change in x can then be thought of as the “average” rate of change of f(x)=x^2 from the first point to the second point. By looking at the graph of f(x)=x^2 it should be clear that it’s “growing” more slowly around (1,1) than it is around (4,16), so if you find an average rate of change near (1,1) vs an average near (4,16) you should get a smaller number.
The ingenious question to ask is, well sure I can find the average rate of change between two points, like between (1,1) and (3,9) by doing the change in y over the change in x. But what about finding the exact rate of change at (3,9) instead? You can try using (3,9) for both points and calculate the average, but you’ll end up dividing by 0. The sneaky trick is to use limits.
So, we find the average rate of change from (3,9) to (3+h, (3+h)^2 ) instead. The idea being that we find the average rate of change from the x-value 3 to a teeny bit more than 3 by adding +h, then later we make h be infinitely small so that 3+h becomes 3.
Finding the average rate of change then gives us:
((3+h)^2 – 9)/(3+h-3)
which becomes with some algebra:
6+h.
Taking the limit as h goes to 0 gives us 6. Then we realize there was nothing special about the point (3,9) and we could instead just use any old point (x,x^2) and do the exact same thing:
((x+h)^2 – x^2 )/(x+h-x)
which again with some algebra becomes:
2x+h
Taking the limit as h goes to 0 gives us 2x. Then we realize there was nothing special about the function f(x)=x^2 and we could do the same process for any other “nice” function:
(f(x+h)-f(x))/(x+h-x)
then take the limit as h goes to 0 to end up with the derivative of f(x), which is another function telling you the rate of change of the original function f(x).
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