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Math Help - Find the slope of the tangent line

  1. #1
    DBA
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    Find the slope of the tangent line

    hi, I did something wrong but I do not what:
    y = f(x)=1/sqrt(x) = x^-1/2

    slope m= lim h->0 {f(x+h) - f(x)]/h

    m=f'(x)= lim h->0 [1/sqrt(x+h) - 1/sqrt(x)]/h

    then I multiply by [sqrt(x+h) + sqrt(x)] / [sqrt(x+h) + sqrt(x)]

    m= lim h->0 {[x-(x+h)]/[sqrt(x+h)*sqrt(x)]} / h

    m= lim h->0 h / [h*(sqrt(x+h)*sqrt(x)] = -1/2sqrt(x)

    This is wrong, it must be -1/ 2sqrt(x^3)

    Thanks for any help
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  2. #2
    Master Of Puppets
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    Quote Originally Posted by DBA View Post
    hi, I did something wrong but I do not what:
    y = f(x)=1/sqrt(x) = x^-1/2
     f(x)= x^{\frac{-1}{2}}

    First principles is very messy here, using y = x^n \Rightarrow y' = nx^{n-1}

     f'(x)= \frac{-1}{2}x^{\frac{-3}{2}}
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  3. #3
    DBA
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    Thank you for the reply, but I need to use the formula to solve the problem
    slope m= lim h->0 [f(x+h) - f(x)]/h

    for f(x) = 1/sqrt(x)

    I know the answer, I just do not know how to get there with this formula.

    Thanks
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