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Math Help - another algebra slip up

  1. #1
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    another algebra slip up

    I have to find the derivative of y=x + 1/x. I understand finding the derivative.

    I split up the two terms and I am using [f(a+h) - f(a)]/h.

    so for the first one it was [(x+h) - (x)]/h = h/h = 1

    the second part was {[1/(x+h)] - [1/x]}/h = {[x/(x(x+h))] - [(x+h)/(x(x+h)]} = {[h/(x(x+h))]}/h then I go to [h/(x(x+h))] * 1/h but from there it messes me up because I get [h/(x^2h +xh^2)]

    I know the final answer is 1-(1/x^2)

    thank you
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  2. #2
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    Re: another algebra slip up

    Quote Originally Posted by Jonroberts74 View Post
    I have to find the derivative of y=x + 1/x. I understand finding the derivative.

    I split up the two terms and I am using [f(a+h) - f(a)]/h.

    so for the first one it was [(x+h) - (x)]/h = h/h = 1

    the second part was {[1/(x+h)] - [1/x]}/h = {[x/(x(x+h))] - [(x+h)/(x(x+h)]} = {[h/(x(x+h))]}/h then I go to [h/(x(x+h))] * 1/h but from there it messes me up because I get [h/(x^2h +xh^2)]

    I know the final answer is 1-(1/x^2)

    thank you
    Let's write this out first:
    \lim_{h \to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h} = \text{ ... } = \lim_{h \to 0} \frac{\frac{h}{x(x + h)}}{h}

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

    Can you finish it now?

    -Dan
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  3. #3
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    Re: another algebra slip up

    then it would be 1h/[x(x+h)h] cancel the h's for 1/x(x+h) then 1/x(x+0) = 1/x^2 ? should be -1/(x^2) thought, right? because of it being 1/x+h) - 1/(x)......
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  4. #4
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    Re: another algebra slip up

    Quote Originally Posted by Jonroberts74 View Post
    then it would be 1h/[x(x+h)h] cancel the h's for 1/x(x+h) then 1/x(x+0) = 1/x^2 ? should be -1/(x^2) thought, right? because of it being 1/x+h) - 1/(x)......
    There is a missing sign above. It should be
    \lim_{h \to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h} = \text{ ... } = \lim_{h \to 0} \frac{\frac{-h}{x(x + h)}}{h}
    Thanks from topsquark and HallsofIvy
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  5. #5
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    Re: another algebra slip up

    thanks, it is very appreciated.
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