Recent content by esrever10

First I got ##f(0)=0##, Then I got ##f'(x)(0)=\frac{\cos x(2+\cosh x)-\sin x\sinh x}{(2+\cosh x)^2}=1/3## But when I tried to got ##f''(x)## and ##f'''(x)##, I felt that's terrible, If there's some easy way to get the anwser?