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Math Help - derivative involving natural log

  1. #1
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    derivative involving natural log

    so the function is y= ln(arcsec(3x+1))

    i know that the derivative of lnx is 1/x

    so, based on this fact, and the chain rule, i said:

    y'= 1/arcsec(3x+1) * 1/x sqrt(x^2-1) * 3

    i know that's not the neatest way to write it.. can someone tell me if this is correct?
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  2. #2
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    Quote Originally Posted by buttonbear View Post
    so the function is y= ln(arcsec(3x+1))

    i know that the derivative of lnx is 1/x

    so, based on this fact, and the chain rule, i said:

    y'= 1/arcsec(3x+1) * 1/x sqrt(x^2-1) * 3

    i know that's not the neatest way to write it.. can someone tell me if this is correct?
    y = \ln{[arcsec{(3x + 1)}]}


    Let u = arcsec{(3x + 1)} so  y = \ln{u}

    \frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}

    \frac{dy}{dx} = u^{-1}\frac{du}{dx}

    \frac{dy}{dx} = [arcsec{(3x+1)}]^{-1}\frac{du}{dx}.


    Now let's find \frac{du}{dx}.

    Let v = 3x + 1 so u = arcsec{\,v}.


    \frac{du}{dx} = \frac{dv}{dx}\,\frac{du}{dv}

    \frac{du}{dx} = 3\frac{1}{v^2\sqrt{1 - \frac{1}{v^2}}}

    \frac{du}{dx} = \frac{3}{(2x + 1)^2\sqrt{1 - \frac{1}{(3x + 1)^2}}}


    So \frac{dy}{dx} = \frac{3}{[arcsec{(3x+1)}](2x + 1)^2\sqrt{1 - \frac{1}{(3x + 1)^2}}}.
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    now i'm really confused

    edit:

    okay, now that i can see the symbols it makes more sense
    i'm not really sure why my method didn't work, though..

    can you just explain where the (2x+1) came from?
    thanks for your help
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    Quote Originally Posted by buttonbear View Post
    now i'm really confused
    Take it one step at a time. You end up using the chain rule twice.

    Edit: Yes, stupid LaTeX doesn't have an inbuilt way of writing arcsec...
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