By using the definition of the derivative, it is possible to find a formula for the derivative function. Recall that the definition of the derivative at some point x is the limit as h approaches 0 of (f(x+h)-f(x))/h.
Substitute the x in f(x) with x+h and evaluate f(x) at this point. By plugging this value into the definition of the derivative and simplifying, you will find a new expression. When h approaches 0, the h's in this expression will become 0, so that the result is solely dependent on x.
We're talking about the derivative function. Here is a problem. Find the formula for the derivative function f'(x) where f(x) is x² plus 1. So the first thing we want to do is recall the definition of the derivative function. F'(x) is the limit as h approaches 0 of f(x+h) minus f(x) over h.
Now usually the first thing I do is I evaluate, and simplify the difference quotient. This inside part of the definition. So f(x) plus h minus f(x) over h for this function. Well, f(x+ h) is going to x plus h quantity squared plus 1. F(x) of course if just x² plus 1. All that over h.
So I have to expand all this, and simplify. There will be a little bit of cancellation. Now x plus h quantity squared is x² plus 2xh plus h², plus1 minus x² minus 12. Don't forget to distribute the minus sign. That's a mistake a lot of people make forgetting to distribute that.
Then you'll notice that a lot of simplification takes place. The x²s cancel, and the 1s cancel. You're left with 2xh plus 8² over h. You'll notice the common factor of h cancels, and you're left with 2x plus h. So you're difference quotients simplifies to 2x plus h.
Now we can pop this value back in for this guys. So we have the limit as h approaches 0 of 2x plus h. That's just 2x. That's our derivative function f'(x) is 2x. The linear function 2x. That's the derivative of f(x) equals x² plus 1.