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Math Help - Differentiating logarithm question

  1. #1
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    Differentiating logarithm question

    I'd really appreciate it if somebody could show me the working for this question. Try as I might I can't seem to get the correct answer to come up. The question is as follows:

    -------------------------------

    By writing the function y = ((2x - 5) / (3x + 2))^0.5 in logarithmic form, show that dy/dx = 19/(2((2x-5)^0.5)((3x + 2)^1.5)

    -------------------------------
    I've used ^ to mean 'to the power of'.

    I've spent a while on this question and I'd really like to understand how it's done, it's annoying me! Please help
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  2. #2
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    Hi !

    Quote Originally Posted by wardj View Post
    I'd really appreciate it if somebody could show me the working for this question. Try as I might I can't seem to get the correct answer to come up. The question is as follows:

    -------------------------------

    By writing the function y = ((2x - 5) / (3x + 2))^0.5 in logarithmic form, show that dy/dx = 19/(2((2x-5)^0.5)((3x + 2)^1.5)

    -------------------------------
    I've used ^ to mean 'to the power of'.

    I've spent a while on this question and I'd really like to understand how it's done, it's annoying me! Please help
    y={\color{red}\frac{2x-5}{(3x+2)^{.5}}}

    Taking the logarithms :

    \ln y=\ln (2x-5)-\ln ((3x+2)^{.5}), because \ln \frac ab=\ln a-\ln b

    Plus, \ln a^n=n \ln a

    Therefore :

    \ln y=\ln (2x-5)-.5 \cdot \ln (3x+2)


    The derivative of \ln y is \frac{\frac{dy}{dx}}{y}=\frac{\frac{dy}{dx}}{\colo  r{red}\dots}

    Can you continue ?
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  3. #3
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    Quote Originally Posted by wardj View Post
    I'd really appreciate it if somebody could show me the working for this question. Try as I might I can't seem to get the correct answer to come up. The question is as follows:

    -------------------------------

    By writing the function y = ((2x - 5) / (3x + 2))^0.5 in logarithmic form, show that dy/dx = 19/(2((2x-5)^0.5)((3x + 2)^1.5)

    -------------------------------
    I've used ^ to mean 'to the power of'.

    I've spent a while on this question and I'd really like to understand how it's done, it's annoying me! Please help

    y=\left( \frac{2x-5}{3x+2}\right)^{\frac{1}{2}}

    Taking the natrual log of both sides gives

    \ln(y)=\ln\left( \frac{2x-5}{3x+2}\right)^{\frac{1}{2}}

    Using log properties we get

    \ln(y)=\frac{1}{2}\ln(2x-5)-\frac{1}{2}\ln(3x+2)

    Now taking the implict derivative of the equation gives

     <br />
\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left( \frac{2}{2x-5} \right)-\frac{1}{2}\left( \frac{3}{3x+2}\right)=\frac{1}{2} \left( \frac{(2(3x+2)-3(2x-5)}{(2x-5)(3x+2)}\right)=

    \frac{1}{y}\frac{dy}{dx}=\frac{1}{2} \left( \frac{19}{(2x-5)(3x+2)}\right) \iff \frac{dy}{dx}=y\frac{1}{2} \left( \frac{19}{(2x-5)(3x+2)}\right)<br />

    Now replace y with what it is equal to and multiply.

    \frac{dy}{dx}=\left( \frac{2x-5}{3x+2}\right)^{\frac{1}{2}}\frac{1}{2} \left( \frac{19}{(2x-5)(3x+2)}\right)=\frac{1}{2} \left( \frac{19}{(2x-5)^{\frac{1}{2}}(3x+2)^{\frac{3}{2}}}\right)

    Yeah
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  4. #4
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    Thanks for the reply, however the top part of the fraction is also square rooted, i.e (2x - 5)^0.5 / (3x + 2)^0.5

    I do get to:

    0.5 ln (2x - 5) - 0.5 ln (3x + 2)

    Even though I have tried to find the derivative with the rule, I can not get to the answer so I would be very grateful if you could outline every step from 0.5 ln (2x - 5) - 0.5 ln (3x + 2) all the way.
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  5. #5
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    Thumbs up

    I see the empty set has already posted!

    Thanks a lot for the help guys, that's exactly what I was looking for. Now to learn how to do it!

    Thanks.
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  6. #6
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    Quote Originally Posted by wardj View Post
    I see the empty set has already posted!

    Thanks a lot for the help guys, that's exactly what I was looking for. Now to learn how to do it!

    Thanks.
    Sorry for misreading your post
    ... It's so rare that people respect like that the parenthesis rules and... I just happen to misread... :'(

    Well, good night
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