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Math Help - second derivitive ..HELP!

  1. #1
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    second derivitive ..HELP!

    f(x)=3ln[sec(x)+tan(x)]
    f(x)=


    I got (((3sec(x))(tan(x))^2)-(3(sec(x))^2)-(2(sec(x))^2)(tan(x))(sec(x)tan(x))+(-3(sec(x)tan(x)+(sec(x))^2)(-sec(x)tan(x)+(sec(x))^2)))/((sec(x)+tan(x))^2)




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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lmao View Post
    f(x)=3ln[sec(x)+tan(x)]
    f(x)=


    I got (((3sec(x))(tan(x))^2)-(3(sec(x))^2)-(2(sec(x))^2)(tan(x))(sec(x)tan(x))+(-3(sec(x)tan(x)+(sec(x))^2)(-sec(x)tan(x)+(sec(x))^2)))/((sec(x)+tan(x))^2)




    you're making your life WWWAAAYYYY too complicated. simplify intermediate steps before moving on

    well, by definition, the derivative of ln|sec(x) + tan(x)| = sec(x), but let's pretend we didn't know that.

    we have to use the chain rule:

    f(x) = 3ln|sec(x) + tan(x)|
    => f ' (x) = 3[1/(sec(x) + tan(x)) * (sec(x)tan(x) + sec^2(x))]
    .............= 3[{sec(x)tan(x) + sec^2(x)}/{sec(x) + tan(x)}]
    .............= 3[sec(x){tan(x) + sec(x)}/{sec(x) + tan(x)}]
    .............= 3sec(x)

    => f '' (x) = 3sec(x)tan(x)
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by lmao View Post
    f(x)=3ln[sec(x)+tan(x)]
    f(x)=


    I got (((3sec(x))(tan(x))^2)-(3(sec(x))^2)-(2(sec(x))^2)(tan(x))(sec(x)tan(x))+(-3(sec(x)tan(x)+(sec(x))^2)(-sec(x)tan(x)+(sec(x))^2)))/((sec(x)+tan(x))^2)




    Well, let's make your life a bit easier:

    f(x) = 3ln[secx+tanx]
    f'(x) = 3*(secxtanx + sec^2x)/(secx + tanx)

    We can simplify the first derivative.
    Factor out secx from the numerator:

    f'(x) = 3secx*(tanx + secx)/(secx + tanx)
    f'(x) = 3sec(x)
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