Question

Can anyone explain derivative for me?

Answer 1

Could you please elaborate on your question? In particular, what is it about derivatives that has you confused?

Answer 2

First, what is definition of derivative?

Answer 3

Do you mean the precise mathematical definition? Or do you want more of an intuitive description of what a derivative is?

Answer 4

Both, please.

Answer 5

First, remember how in algebra you learned to calculate the slope of a straight line? You just take the "rise" and divide it by the "run"--i.e. you choose two distinct points on the line and calculate the change in y and divide it by the change in x.
Now, when you're calculating this slope, you can choose any two distinct points on the line because the slope of a straight line is constant--it is the same everywhere. However, this only works for straight lines. What do we do when we have curves that aren't straight? Well, this is where Calculus comes in. In particular, we calculate the derivative of a function in order to determine the exact slope of a curve at any particular point we want.
Are you with me so far??

Answer 6

Yes, I am. Thank you so much!
Now please tell me how to find the derivative of a function all right?

Answer 7

Ok, so we calculate the derivative of a function "f(x)" with respect to x as:
\[\ = \lim_{h \rightarrow 0} \frac{ f(x +h) - f(x)}{ h }\]
Two important things are going on:
First, you are calculating the slope (like you normally would do with a straight line) between the points (x, f(x)) and the point (x+h, f(x+h)). [Note: you should think of the "h" term as representing some small amount.]
Second, by taking the "limit" of this slope as "h" goes to zero, you are making "h" as arbitrarily small as you like so that h gets closer and closer (but never equal to) zero. [Note: we say that "h" gets arbitrarily close to zero without ever actually equalling zero so that we avoid ever having to divide by zero--remember, dividing by "0" is a HUGE "NO!" in math...]
So this is the technical definition of the derivative, now give me a sec and I'll give you an example...

Answer 8

First, I want to clarify what I mean by "f(x+h)". Let's say you start out with a function such as:
\[f(x) = x ^{2} + x -7\]
Now, when I write "f(x+h)", what that means is that wherever you see "x" in the previous formula, you replace it with "x+h", giving you:
\[f(x+h) = (x+h)^{2} + (x+h) -7\] Which can be reduced (via algebra) to:
\[= x ^{2} + 2xh +h ^{2} +x + h -7\]

Answer 9

Ok, so back to the derivative... So, the derivative of "f" with respect to x (often written as "f'(x)"] is given by:
\[f'(x) = \lim_{h \rightarrow 0}\frac{ f(x+h)-f(x) }{ h }\]
For the function:
\[f(x) = x ^{2}+x-7\] The derivative is written as:
\[f'(x) = \lim_{h \rightarrow 0}\frac{ (x+h) ^{2}+(x+h)-7) - (x ^{2}+x-7) }{ h }\]
Which simplifies to:
\[= \lim_{h \rightarrow 0} \frac{ 2xh +h + h ^{2} }{ h }\]
Then, we factor out an "h" from the numerator which then cancels out the "h" in the denominator, leaving us with:
\[=\lim_{h \rightarrow 0} 2x+1 + h \]
And when limit as h goes to zero of that last expression is just "2x+1".
Thus,
\[f \prime(x) = 2x+1\]
So, when x=3, for instance, the value of f is "5" and the value of the derivative at x=3 is "7". Does this make sense?
Actually, if you want a better explanation you might want to check out the videos over at "Khan Academy" such as this one:

https://www.khanacademy.org/math/calculus/differential-calculus/derivative_intro/v/calculus--derivatives-1--new-hd-version

Hope this is helpful!