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Math Help - irritating derivative problem

  1. #1
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    Angry irritating derivative problem [solved]

    hi, i have a test on derivatives tomorrow and was going over some problems when I got stumped. I have the problem and the answer, but i want to know how to get the answer. thanks

    use definition fprime (a) = (lim as h approaches 0) [f(a+h) - f(a)] / h

    to find the derivative of the given function at the indicated point

    f(x) = 1/x, a = 2

    nvm, brain cramp, i figured it out
    Last edited by TacticalPro; October 19th 2011 at 05:21 PM.
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  2. #2
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    Re: irritating derivative problem

    f'(a) = \lim_{h \to 0} \frac{\frac{1}{a+h} - \frac{1}{a}}{h}

    = \lim_{h \to 0} \frac{\frac{a}{a(a+h)} - \frac{a+h}{a(a+h)}}{h}

    = \lim_{h \to 0}\left(\frac{1}{h}\right)\left(\frac{-h}{a(a+h)}\right)

    = \lim_{h \to 0}\frac{-1}{a(a+h)}

    = \frac{-1}{a^2}, provided a ≠ 0.

    since 2 ≠ 0, f'(2) = -1/4.
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  3. #3
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    Re: irritating derivative problem

    Quote Originally Posted by TacticalPro View Post
    hi, i have a test on derivatives tomorrow and was going over some problems when I got stumped. I have the problem and the answer, but i want to know how to get the answer. thanks

    use definition fprime (a) = (lim as h approaches 0) [f(a+h) - f(a)] / h

    to find the derivative of the given function at the indicated point

    f(x) = 1/x, a = 2
    \lim_{h \to 0} \frac{1}{h} \left(\frac{1}{2+h} - \frac{1}{2}\right)

    get a common denominator and combine the fractions in ( ) ... do it right and you'll get that h in the denominator to cancel out ... then determine the limit.
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  4. #4
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    Re: irritating derivative problem

    Quote Originally Posted by skeeter View Post
    \lim_{h \to 0} \frac{1}{h} \left(\frac{1}{2+h} - \frac{1}{2}\right)

    get a common denominator and combine the fractions in ( ) ... do it right and you'll get that h in the denominator to cancel out ... then determine the limit.
    thanks, when i first looked at the problem i completely forgot about rearranging to get the h out of the botom
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  5. #5
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    Re: irritating derivative problem

    Quote Originally Posted by Deveno View Post
    \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}

    = \lim_{h \to 0} \frac{\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)}}{h}

    = \lim_{h \to 0}\left(\frac{1}{h}\right)\left(\frac{-h}{x(x+h)}\right)

    = \lim_{h \to 0}\frac{-1}{x(x+h)}

    = \frac{-1}{x^2}, provided x ≠ 0.

    since 2 ≠ 0, f'(2) = -1/4.
    thanks!!!! I actually plugged in for x before rearranging and I find I like it your way better: rearranging the equation in terms of x, and then putting in 2 for x to get the derivative. !!! yay
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