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

1 Answer
Mar 14, 2018

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

Explanation:

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