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Math Help - logorithmic differentiation

  1. #1
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    logorithmic differentiation



    Heres what i did:

    used the laws of logs to get this:

    7x ln (sqrt(x))

    ln(y) = 7x ln (sqrt(x))

    start solving by implicit differentiation:

    y'/y = 7x*(1/x) *(1/2sqrt(x))+ln(sqrt(x))*7

    final = y'=y(7x*(1/x) *(1/2sqrt(x))+ln(sqrt(x))*7)

    I think i know what i did wrong, i used ln(sqrt(x)) as 1 function instead of 2. This is a Web Assign and im about out of guesses. If I would apply the product rule between 7x and ln then apply it in another product rule with sqrt(x) would i get the right answer?
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  2. #2
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    Quote Originally Posted by Evan.Kimia View Post


    Heres what i did:

    used the laws of logs to get this:

    7x ln (sqrt(x))

    ln(y) = 7x ln (sqrt(x))

    start solving by implicit differentiation:

    y'/y = 7x*(1/x) *(1/2sqrt(x))+ln(sqrt(x))*7

    final = y'=y(7x*(1/x) *(1/2sqrt(x))+ln(sqrt(x))*7)

    I think i know what i did wrong, i used ln(sqrt(x)) as 1 function instead of 2. This is a Web Assign and im about out of guesses. If I would apply the product rule between 7x and ln then apply it in another product rule with sqrt(x) would i get the right answer?

    the equation can be further simpilifed before you take the derivative:

    \ln(y)=7x\ln(\sqrt{x})=\frac{7x}{2}\ln(x)

    You could also note this from the beginning

    y=(\sqrt{x})^{7x}=(x^{1/2})^{7x}=x^{\frac{7x}{2}}

    Then take the log

    So then we get

    \frac{y'}{y}=\frac{7}{2}\ln(x)+\frac{7}{2}= ...

    I hope this helps
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