# Math Help - Trig derivative question

1. ## 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....

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