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Math Help - Trig derivative question

  1. #1
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    Trig derivative question

    The second derivative of -(cosx)ln(secx+tanx)

    Well for the first derivative I got sinx*ln(secx+tanx)-(cosx*(secxtanx+sec^2x)/(secx+tanx))....have I missed out some simplification coz this looks ugly....
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  2. #2
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    Quote Originally Posted by daskywalker View Post
    The second derivative of -(cosx)ln(secx+tanx)

    Well for the first derivative I got sinx*ln(secx+tanx)-(cosx*(secxtanx+sec^2x)/(secx+tanx))....have I missed out some simplification coz this looks ugly....

    \sec x+\tan x=\frac{\sin x + 1}{\cos x}\Longrightarrow \ln(\sec x+\tan x)=\frac{\cos x}{\sin x+1}\cdot\frac{\cos^2x+\sin^2x+sinx}{\cos^2x} =\frac{1}{\cos x}=\sec x , so:

    [-\cos x\ln(\sec x+\tan x)]'=\sin x\ln(\sec x+\tan x)-1

    Now I leave the 2nd derivative to you...since you alread saw the 1st one is not that horrendous.

    Tonio
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