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)))'=((\sin(2x))')/(\sin(2x))=(\cos(2x)(2x)')/(\sin(2x))=(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(\frac\pi4\right)=-4\csc^2\left(\frac{2\pi}4\right)=-4`