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Math Help - Simple question -definition of derivative

  1. #1
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    Simple question -definition of derivative

    We've all seen the definition, but for some reason I still don't understand the logic behind it. This is the part of the definition that I have a problem with:

    f (x+h) - f (x) / (x+h) - x

    I understand that (x+h) - x = h.

    If this is the case, will this be true f (x+h) - f (x) = h ...?

    Thus, h / h ?
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  2. #2
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    Hi there,

    This part is given

    \frac{f (x+h) - f (x)}{(x+h) - x}

    then as you have shown there is some cancellation in the denominator

    \frac{f (x+h) - f (x)}{h}

    This is where it stops. No more work needs to be done.

    If you want to find a derivative from here find the limit of your function as h \to 0 using the deifnition above.
    Last edited by pickslides; March 7th 2010 at 02:49 PM. Reason: Bad Latex
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  3. #3
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    Quote Originally Posted by pinkperil View Post
    We've all seen the definition, but for some reason I still don't understand the logic behind it.
    Do you understand how to determinte the slope of the chord between the two points \left(x,f(x)\right)~\&~\left(x+h,f(x+h)\right)?
    That is the logic of it all.
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  4. #4
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    Quote Originally Posted by pinkperil View Post
    We've all seen the definition, but for some reason I still don't understand the logic behind it. This is the part of the definition that I have a problem with:

    f (x+h) - f (x) / (x+h) - x

    I understand that (x+h) - x = h.

    If this is the case, will this be true f (x+h) - f (x) = h ...?
    NO, f(x+ h) is NOT equal to f(x)+ h! Exactly what it is depends on the function f.

    Thus, h / h ?
    Well, that would make calculus pretty trivial wouldn't it?

    If f(x) is a linear function, f(x)= ax+ b, then f(x+h)= a(x+h)+ b= ax+ah+ b.
    f(x+h)= ax+ah+ b- ax+b= ah. (f(x+h)- f(x))/h= ah/h= a, the slope of the line.

    If f(x)= ax^2+ bx+ c, then f(x+h)= a(x+h)^2+ b(x+h+ c = ax^2+ 2axh+ h^2+ bx+bh+ c so f(x+h)- f(x)= ax^2+ 2axh+ h^2+ bx+ bh+ c- ax^2- bx- c= 2axh+ h^2+ bh and the "difference quotient" \frac{f(x+h)- f(x)}{h}= \frac{2axh+ h^2+ bh}{h}= 2ax+ h+ b

    It gets really complicted for things like sin(x), cos(x), or log(x)!
    Last edited by HallsofIvy; March 8th 2010 at 04:54 AM.
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    Quote Originally Posted by HallsofIvy View Post
    NO, f(x+ h) is NOT equal to f(x)+ h! Exactly what it is depends on the function f.
    I see where my problem is. My problem is something even more fundamental: misunderstanding of functions. So if the function isn't what I assumed, then I must assume this:

    y = f(x) = 9

    f(x+h) - f(x) / h

    9 - 9 / h

    Although, it doesn't make much sense to me how f(x + h) = f(x) when this isn't true in many other cases like this:

    f(x) = 2x
    f(x + h) = 2x + 2h

    Is it because the first situation has only a constant and the other has a variable? The reason why I thought f(x+h) - f(x) = h is because: (constant + h) - constant = h.
    Last edited by pinkperil; March 8th 2010 at 01:23 AM.
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    Quote Originally Posted by pinkperil View Post
    I see where my problem is. My problem is something even more fundamental: misunderstanding of functions. So if the function isn't what I assumed, then I must assume this:

    y = f(x) = 9

    f(x+h) - f(x) / h

    9 - 9 / h

    Although, it doesn't make much sense to me how f(x + h) = f(x) when this isn't true in many other cases like this:
    You quoted part of my post so you must have read it. How could you get "f(x+h)= f(x)" from it? f(x+h)= f(x) only for a constant function such as "f(x)= 9" where f(x)= f(y) no matter what x and y are.

    And why would you say "if the function isn't what I assumed". You did not mention any particular function in your original post.

    f(x) = 2x
    f(x + h) = 2x + 2h

    Is it because the first situation has only a constant and the other has a variable? The reason why I thought f(x+h) - f(x) = h is because: (constant + h) - constant = h.
    If f is a constant, f(x)= c, then f(x+h)= c also so f(x+h)- f(x)= 0 and then (f(x+h)- f(x))/h= 0. Of course, the graph of a constant function is a horizontal straight line and has slope 0.

    The function f(x)= x+ c has f(x+h)= x+ h+ c so f(x+ c)- f(x)= x+h+c- x- c= h and then (f(x+h)- f(x))/h= 1. The line y= x+ c has slope 1.

    IF you were refering to specific function you should have said so in your first post. It appeared that your "f(x)" could be any function.
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  7. #7
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    ...only for a constant function such as "f(x)= 9" where f(x)= f(y) no matter what x and y are.
    This makes sense to me now. I couldn't wrap my head around (x + h) being something whole and my mind kept wanting to break it apart. Thanks a lot for your help. I appreciate it.
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