Use the chain rule to compute the second derivative:
[tex]f(x)=\ln(\sin(2x))[/tex]
The first derivative is
[tex]f'(x)=(\ln(\sin(2x)))'=\dfrac{(\sin(2x))'}{\sin(2x)}=\dfrac{\cos(2x)(2x)'}{\sin(2x)}=\dfrac{2\cos(2x)}{\sin(2x)}[/tex]
[tex]f'(x)=2\cot(2x)[/tex]
Then the second derivative is
[tex]f''(x)=(2\cot(2x))'=-2\csc^2(2x)(2x)'[/tex]
[tex]f''(x)=-4\csc^2(2x)[/tex]
Then plug in π/4 for x :
[tex]f''\left(\dfrac\pi4\right)=-4\csc^2\left(\dfrac{2\pi}4\right)=-4[/tex]
