Question

If f(x)=ln(sin(2x)), f''(π/4) is equal to​

Answer 1

Use the chain rule to compute the second derivative:

$f(x)=\ln(\sin(2x))$

The first derivative is

$f'(x)=(\ln(\sin(2x)))'=\dfrac{(\sin(2x))'}{\sin(2x)}=\dfrac{\cos(2x)(2x)'}{\sin(2x)}=\dfrac{2\cos(2x)}{\sin(2x)}$

$f'(x)=2\cot(2x)$

Then the second derivative is

$f''(x)=(2\cot(2x))'=-2\csc^2(2x)(2x)'$

$f''(x)=-4\csc^2(2x)$

Then plug in π/4 for x :

$f''\left(\dfrac\pi4\right)=-4\csc^2\left(\dfrac{2\pi}4\right)=-4$