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Math Help - first, second and third deriative of ln (tan x )

  1. #1
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    first, second and third deriative of ln (tan x )

    y=ln(tan x), dy/dx=2/(sin2x) and d2y/dx2= -4(cos 2x)/(sin 2x)^2 , show that d3y/dx3 = ((4)(3+cos 4x))/(sin 2x)^3 ... i got the solution for dy/dx and d2y/dx2 but not d3y/dx3 , can anyone show me how to get d3y/dx3 please? Thanks in advance!

    here's the working
    https://www.flickr.com/photos/123101...3/13824984513/
    Last edited by delso; April 13th 2014 at 08:23 AM.
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  2. #2
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    Re: first, second and third deriative of ln (tan x )

    Hello, delso!

    y\:=\:\ln(\tan x).\;\;\;\text{Find }\,\frac{dy}{dx},\;\frac{d^2y}{dx^2},\;\frac{d^3y}  {dx^3}

    I would do it this way . . .


    y \;=\;\ln(\tan x)


    \frac{dy}{dx} \;=\;\frac{1}{\tan x}\sec^2x \;=\;\frac{\cos x}{\sin x}\cdot\frac{1}{\cos^2x} \;=\;\frac{1}{\sin x\cos x}

    . . =\;\frac{2}{2\sin x\cos x} \;=\;\frac{2}{\sin2x} \;=\;2\csc2x


    \frac{d^2y}{dx^2} \;=\;2\cdot(-\csc2x\cot2x)\cdot2 \;=\;-4\csc2x\cot2x


    \frac{d^3y}{dx^3} \;=\;-4\bigg[\csc2x(-\csc^22x\cdot 2) + \cot2x(-\csc2x\cot2x\cdot 2)\bigg]

    . . . =\;8\:\!\csc2x(\csc^22x + \cot^22x)
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  3. #3
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    Re: first, second and third deriative of ln (tan x )

    can you show that d3y/dx3 = ((4)(3+cos 4x))/(sin 2x)^3 ? i'm quite weak when it comes to trigonometry....thanks in advance!
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  4. #4
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    Re: first, second and third deriative of ln (tan x )

    Start by replacing \cos{4x} with \cos^2{2x}-\sin^2{2x} and 3 with 3\cos^2{2x}+3\sin^2{2x}.

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