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    Hey Guys, another question...

    if you had to differentiate ln3x you would assume you could do it like this:

    ln3x = ln3 + lnx

    so you will just get 1/x as ln3 is a constant

    Howevere... If you did the chain rule and made u = 3x you would get 3/x as your answer, but this is wrong.

    However some questions like - differentiate ln (3x + 1) you have to use the chain rule, I don't understand how you can use the chain rule here and get a correct answer, but if you used in the first instance you get the wrong answer.

    Thanks
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    y=ln(u) where u=3x
    \frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}
    =\frac{1}{u} \cdot 3
    =3/3x , 3's cancel
    =1/x
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    You can differentiate ln(3x) different ways. Each way will produce 1/x and some constant. The constants are all different however so you actually haven't done it wrong.
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    (Original post by eddhann)
    However some questions like - differentiate ln (3x + 1) you have to use the chain rule
    ln(3x+1)=ln(3(x+1/3)) is just a translation of ln(3x), so you can work out the derivative of ln(3x) and translate it rather than using the chain rule. It's probably easier to use the chain rule, but you don't have to.
    (Original post by jaheen22)
    You can differentiate ln(3x) different ways. Each way will produce 1/x and some constant. The constants are all different however so you actually haven't done it wrong.
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    (Original post by eddhann)
    Hey Guys, another question...

    if you had to differentiate ln3x you would assume you could do it like this:

    ln3x = ln3 + lnx

    so you will just get 1/x as ln3 is a constant

    Howevere... If you did the chain rule and made u = 3x you would get 3/x as your answer, but this is wrong.

    However some questions like - differentiate ln (3x + 1) you have to use the chain rule, I don't understand how you can use the chain rule here and get a correct answer, but if you used in the first instance you get the wrong answer.

    Thanks
    It seems you're not understanding the chain rule. You're correct in saying that ln3x = ln3 + lnx and hence \frac{d}{dx}ln3x=\frac{1}{x}.

    In order to do the chain rule, do the following: \text{let } y = ln3x, u = 3x\Rightarrow y = lnu
    Then differentiate according to the chain rule:

    \frac{du}{dx}=3, \frac{dy}{du}=\frac{1}{u}
    \frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}=3\cdot \frac{1}{u}=\frac{3}{u}=\frac{3}  {3x}=\frac{1}{x}
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    For this we can say 3x=f(x). Now for ln (x) you can basically say dy/dx = f`(x)/f(x). So dy/dx= 1/x.

    Or for using the chain rule we can say this let u=3x du/dx=3 and then say y=ln(u) so dy/du = 1/u. Now since dy/dx = du/dx * dy/du we can say dy/dx= 3/u =3/3x =1/x.
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    (Original post by harr)
    ?
    I'm on drugs :cool:
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    thanks everyone
 
 
 
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