Help me understanding about derivatives???

Question

anonymous · Answer

ok
I'll try

anonymous · Answer

a specific question is best
if you want a detailed explanation is text book or google search will be better than asking a general question here

mandre · Answer

Derivatives can get quite intricate. It depends a lot on how much you already understand. If you're doing it in school your best bet would be reading your textbook and/or asking your teacher / lecturer or fellow students.

mandre · Answer

Or since you're online, you can Google it and find pages about derivatives for all levels.

phi · Answer

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

theeric · Answer

I've heard that khanacademy is great! So definitely check that out! And internet searches can help! If you like informal, quick textbook style lessons, you should also check out Paul's Online Math Notes. Here is the part for calculus: http://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx . I'm not the best at math, so I'm hoping somebody can correct me as necessary. Derivatives.... Well, the MAIN thing is that they take a function, and they can find the rate of change of that function "with respect to" the independent variable. "With respect to" is just the common way to say it. It makes sense, though. It means that you want to see how the function changes due to only the change of the independent variable. The rate of change is very easy to understand. It is the change in the function value per the change in the independent variable value. Note that I mean CHANGE. That's important. Many people use the standard $y$ is a function of $x$, often denoted as $y=f(x)$ where $f$ is a function. Then the slope is the change in $y$ per the change in $x$. If you're familiar with point-slope formula and other basic line equations (it's okay if they're fuzzy), then you probably realize that this is the slope of a line! For a curve, it's not so easy to find slope at its points. Ordinarily, you could only estimate this on a curve. Until calculus. Here's the general attempt to approximate the slope at a point. You go to that area of the curve and pick two points. You find the change in $y$ and change in $x$. Then you do the typical "rise over run" thing. I'll use $\Delta$ to show the change in that variable (it's what is commonly done). So the rate of change of $y$ with respect to $x$ (as $x$ changes) is $\dfrac{\Delta y}{\Delta x}$.|dw:1377239005867:dw| But, that is just an estimate. If it was exact, the line would exactly fit the angle of the curve at the point. You can see it doesn't in the approximation. So, now we use calculus! So, you might see that when the two points we choose are closer, the slope of the line through them is closer to the slope of the curve!|dw:1377239586132:dw||dw:1377239700865:dw| Now what if we could make it so close.... So close it is just that point? The second point is the first point. Makes no sense, right? Right. But we can do the next best thing. The point IMMEDIATELY next to the first point. This is where the study of limits is important. They teach you that you can use some rules to find the value of a function as the variable gets very close to a value. So, derivatives focus on the change in $x$ being $x+h|), where \(h$ is just a amount added to $x$, minus $x$. So it is the change in $x$.|dw:1377240128764:dw| So the $\Delta x = (x+h)-x=h$. And the change in a function of $x$ would be $\Delta y=f(x+h)-f(x)$. The slope is $\dfrac{\Delta y}{\Delta x}$, remember. So the slope is $\dfrac{f(x+h)-f(x)}{h}$. The slope at $x,\ f(x)$ is most accurate as $h$ gets closer to $0$.