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Math Help - [SOLVED] Help finding f(x)'.... THX!

  1. #1
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    [SOLVED] Help finding f(x)'.... THX!

    Can someone give me a hint with this problem?
    The problem is:

    (tan x)^(arctan x)

    I was trying to use:
    Dx (a)^u = (a)^u * ln(a) * Dx u

    I am getting:
    (tan)^(arctan x) * ln(tanx) * 1/(1+x^2)

    The answer in the back of my book is:
    (tan)^(arctan x) * [arctan(x)cot(x)sec^2(x) + ln(tanx)/(1+x^2)

    I am messing up with the chain rule somewhere I think =(.

    Again just a hint will do! Thanks!!!
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  2. #2
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    Dear stickyMATH,

    f(x)=(tanx)^{arctanx}

    lnf(x)=arc(tanx).ln(tanx)

    Differenciating w.r.t x,

    \frac{d}{df(x)}lnf(x)\frac{df(x)}{dx}=ln(tanx)\fra  c{d}{dx}(arctanx)+arctanx\frac{d}{dx}ln(tanx)

    \frac{1}{f(x)}\frac{df(x)}{dx}=\frac{ln(tanx)}{1+x  ^2}+(arctanx)cotxsec^{2}x

    \frac{df(x)}{dx}=(tanx)^{arctan x}\left[(arctanx)cot(x)sec^2(x) + \frac{ln(tanx)}{(1+x^2)}\right]<br />

    Hope this helps.
    Last edited by Sudharaka; January 29th 2010 at 07:33 PM.
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