I am in highschool calculus, we have just started the unit on derivatives and I cannot figure out how to do this one question:
how would I find the f'(x) using the (f(x+h)f(x))/H method.

I am in highschool calculus, we have just started the unit on derivatives and I cannot figure out how to do this one question:
how would I find the f'(x) using the (f(x+h)f(x))/H method.
Why don't you show us what you have done.Originally Posted by Gmano
No one is going to do your school work for you, but perhaps we can ask some questions that will help you to solve the problem yourself.
Try using the quotient rule. It states that if you have a function , than its derivative, that is , is equal to .Originally Posted by Gmano
He's gotta use the f(x+h) limit method. Product/quotient rules haven't been introduced yet except for the fact that the limit of a quotient is the quotient of the limits (hint! hint!).
Oh, I see. Thank you Ots.
Gmano,
You should recall the difference quotient is more tedious than using the rules to differentiate, and it's likely your teacher will soon move past this method (after you have proven to be able to differentiate [or your class has] functions using the "definition," so to speak).
Here is the function put into the difference quotient:
You're going to simplify the above expression such that the output as the limit approaches 0 is not undefined, and that will be your derivative.
They have learned at this point to treat the numerator and denominator seperately and then do a quotient on the respective limits after they have been taken. Much easier. That's what I meant by the limit of a quotient is the quotient of the limits. But I think he may have moved on by now.
\Originally Posted by Ots
that doesn't work for derivatives. In every case you find your limit using that approach is 0/0 which is completely meaningless.
The limit of a quotient is NOT the quotient of the limits when the limit of the denominator is 0. That is ALWAYS the case when looking at the quotient that defines the derivative. That is the whole poiint of the derivative.
Yes, it only works for limits. And when the limit of the denominator is not zero. So, he must do the problem using the quotient rule as Ellatha pointed out but with the f(x+h) method of taking the derivative.
No he doesn't have to use the quotient rule. In fact the whole point of his exercise is to show what the derivative is from the basic definition without using the quotient rule or any other ruleOriginally Posted by Ots
Eliatha in the second point set up the problem properly.
The quotient rule is nothing more than a rule for taking the derivative of a quotient of functions. It is derivable from the basic definition of the derivative, and is not magic.
The point of this exercise is to use the definition directly. You seem to have missed that point, among others.
One might prove the quotient rule using the definition. But that is above and beyond the requirements of the original problem.
Note to Gmano. You would do well to igore Ots. He doesn't know what he is talking about. He rarely does.
Nope. But at least your record remains umblemished (by success).Originally Posted by Ots
So, the quotient and other rules are essentially a shorthand way of performing the calculations of the formal limit definiton. Is that right?
You need the formal definition and the geometric intuition that goes with it to understand what a derivative is and why anyone would want to compute one.Originally Posted by CMR80606
Once you understand that, then one want to simplify the process of actually computing derivatives for the common functions that one encounters in applications. (You get to the point where you can do these things in your head, whch allows you to focus on the physics or engineering at hand and not on just computing the derivative).
The various "rules" like th quotient rule, the product rule and the chain rule, allow one to compute the derivative efficiently. You will also find later that they play a role in integration, and a theoretical role in more advanced applications of calculus.
Once you are a little farther on, you will find that it is rare indeed to calculate the value of a derivative directly from the definition. But it is good exercise while you are learning the subject and helps you to appreciate what a derivative really is.
There is a bit more on the general idea behind limits and derivatives here.
http://www.thescienceforum.com/limit...ion21311t.php
« Annual temperature range  is phi real? » 