What would be the second order derivative of the following function? #x^2log|cosx|#

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Answer 1 · 2018-03-14T18:19:39

If I'm not mistaken the answer should be #2ln(cos(x))-4xtg(x)+x^2/cos^2(x)#

So #f''(x)=?# with #f(x)=x^2ln(cos(x))#

#f'(x)=d/dx(x^2ln(cos(x)))#

#=d/dxx^2*ln(cos(x))+d/dx(ln(cos(x))*x^2#

#=2xln(cos(x))+x^2*1/cos(x)*d/dx(cos(x) )#

#=2xln(cos(x))+x^2*(-sin(x)/cos(x) )#

#=2xln(cos(x))-x^2tg(x) = f'(x)#

#f''(x)=d/dx(2xln(cos(x))-x^2tg(x))#

#=2ln(cos(x))+2xd/dxln(cos(x))-2xtg(x)+x^2*d/dxtg(x)#

#=2ln(cos(x))-2xtg(x)-2xtg(x)+x^2*(1/cos^2(x))#

#=2ln(cos(x))-4xtg(x)+(x/cos(x))^2#