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Math Help - derivative of 1/ln

  1. #1
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    derivative of 1/ln

    Hey guys, I can't seem to figure this problem out, can anyone help??
    its:

    y = 1 / ln (cx)

    find derivate of y... and the value of c is found by rearranging the equation above and isolating c

    I keep getting that c = 1 / xe^y
    and then i get that the derivative is y' = e^y / (ln (1/e^y)^2 but i know its wrong because the answer should be: y' = -y^2 / x
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  2. #2
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    Quote Originally Posted by collegestudent321 View Post
    Hey guys, I can't seem to figure this problem out, can anyone help??
    its:

    y = 1 / ln (cx)

    find derivate of y... and the value of c is found by rearranging the equation above and isolating c

    I keep getting that c = 1 / xe^y
    and then i get that the derivative is y' = e^y / (ln (1/e^y)^2 but i know its wrong because the answer should be: y' = -y^2 / x
    y = [\ln(cx)]^{-1}

    y' = -[\ln(cx)]^{-2} \cdot \frac{1}{x}

    y' = -\left([\ln(cx)]^{-1}\right)^2 \cdot \frac{1}{x}

    y' = -(y)^2 \cdot \frac{1}{x} = -\frac{y^2}{x}
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  3. #3
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    Quote Originally Posted by collegestudent321 View Post
    Hey guys, I can't seem to figure this problem out, can anyone help??
    its:

    y = 1 / ln (cx)

    find derivate of y... and the value of c is found by rearranging the equation above and isolating c

    I keep getting that c = 1 / xe^y
    and then i get that the derivative is y' = e^y / (ln (1/e^y)^2 but i know its wrong because the answer should be: y' = -y^2 / x
    do you know implicit differentiation ? if you do write as

    yln(cx)=1

    if not by quotient rule with

    u=1 du=0
    v=ln(cx) dv=1/x

     <br />
y'= \frac {(\frac{-1}{x})}{(ln(cx))^2}<br />

    tidy this up and use

     <br />
y=\frac{1}{ ln(cx)}<br />

    to get the answer
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  4. #4
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    Quote Originally Posted by collegestudent321 View Post
    Hey guys, I can't seem to figure this problem out, can anyone help??
    its:

    y = 1 / ln (cx)

    find derivate of y... and the value of c is found by rearranging the equation above and isolating c

    I keep getting that c = 1 / xe^y
    and then i get that the derivative is y' = e^y / (ln (1/e^y)^2 but i know its wrong because the answer should be: y' = -y^2 / x
    y=(\ln{(cx)})^{-1}

    y'=-(\ln{(cx)})^{-2}(\frac{1}{cx})(c)

    y'=-(y)^2\frac{1}{x}
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  5. #5
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    got it! makes perfect sense! thanx guys!
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