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A few quick questions with derivatives using limit def.

  1. Mar 28, 2008 #1
    1. The problem statement, all variables and given/known data
    The first one says use the limit definition to find the derivative of the function.
    F(x)= 1/(2x-4)

    the second one is use the limit definition to find the derivative at the indicated point..
    f(x)= -x^3 + 4x^2, at (-1,5)

    the last one is use the limit process to find the slope of the graph of the function at the specified point.
    f(x)= sqrt of (x + 10), at (-1,3)

    2. Relevant equations
    i know that this equation is used to help find it..
    F(x + P)- f(x)/ P
    p=the change in x..or delta x

    3. The attempt at a solution
    i dont even know where to begin ...but any help at all is welcomed and thanked.
  2. jcsd
  3. Mar 28, 2008 #2
    Well, first of all the definition of the derivative, as you may know is:

    [tex]f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h},f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}[/tex]

    For your first function it would be

    [tex]f'(x)=\lim_{h\rightarrow 0}\frac{\frac{1}{2(x+h)-4}-\frac{1}{2x-4}}{h}[/tex] Now you can go from here, right?

    Second one:

    [tex]f'(-1)=\lim_{x\rightarrow -1}\frac{-x^{3}+4x^{2} -f(-1)}{x-(-1)}[/tex]

    Third one

    [tex]slope=m=f'(-1)=\lim_{x\rightarrow -1}\frac{\sqrt{x+10}-\sqrt{-1+10}}{x-(-1)}[/tex]

    Now all you need to do is evaluate those limits. DO u know how to do it?
  4. Mar 28, 2008 #3
    Blah, i forgot: You NEED to show your work next time, before anyone here can help you.
  5. Mar 28, 2008 #4
    For example, if I wanted to find the derivative of the function 2x^2 + 5 using the definition of a limit, I would use the formula:

    [tex]\lim_{\substack{h\rightarrow 0}}\frac{f(x+h)-f(x)}{h}=\lim_{\substack{h\rightarrow 0}}\frac{2(x+h)^2+5-(2x^2+5)}{h}[/tex]

    [tex]=\lim_{\substack{h\rightarrow 0}}\frac{2x^2+4xh+2h^2+5-2x^2-5}{h}=\lim_{\substack{h\rightarrow 0}}\frac{4xh+2h^2}{h}[/tex]

    [tex]=\lim_{\substack{h\rightarrow 0}}4x+2h=4x[/tex]

    See how f(x) = 2x^2 + 5 and f(x+h) = 2(x+h)^2 + 5?
    Last edited: Mar 28, 2008
  6. Mar 29, 2008 #5
    To the OP:In order to recieve further help, please show us what you have tried so far, and point out where are u stuck. Remember you are supposed to do your own homework not us.!!
  7. Mar 29, 2008 #6
    ok i didnt know that
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