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Math Help - Maclaurin's Series question

  1. #1
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    Maclaurin's Series question

    I can't get this:

    y = ln(1 + tan x)

    (i) Show that d2y/dx2 = -1 when x = 0.

    Thanks if you could help, I'm preparing for my promos next week!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by margaritas View Post
    I can't get this:

    y = ln(1 + tan x)
    Step 1 use the chain rule:

    dy/dx=[1/(1+tan(x))](sec(x))^2

    Step 2 use product rule (you could use the quotient ruleif you prefer)
    and chain rule:

    d^2y/dx^2=[1/(1+tan(x))^2](-1) [sec(x)]^4 + [1/(1+tan(x))].2.[sec(x)]^2 tan(x)

    (i) Show that d2y/dx2 = -1 when x = 0.
    when x=0 tan(x)=0, and sec(x)=1, result follows.

    RonL
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  3. #3
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    Hello, margaritas!

    It's just two applications of the Quotient Rule . . .
    . . Exactly where is your difficulty?


    y = ln(1 + tan x)

    (i) Show that dy/dx = -1 when x = 0

    dy . . . . . .secx
    --- . = . ------------
    dx . . . . 1 + tan x


    dy . . . (1 + tan x)2sec x(sec xtan x) - secxsecx
    ---- .= .------------------------------------------------------
    dx . . . . . . . . . . . . . (1 + tan x)

    . . . . . . 2secxtan x(1 + tan x) - secxsecx
    . . . = . ----------------------------------------------
    . . . . . . . . . . . . . (1 + tan x)


    When x = 0, we have:

    . . . dy . . . . 2sec0tan 0(1 + tan 0) - sec0sec0
    . . . ---- . = . ----------------------------------------------- . = . -1
    . . . dx . . . . . . . . . . . (1 + tan 0)

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  4. #4
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    Oh right I got it! I was probably put off by the fact that it was so tedious to convert sec^2 x to 1/cos^2x and then converting further again using the double angle formula, but anyway thanks guys for your help!
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