Question
If $$f\left( x \right) = \cos \,x.\cos \,2x.\cos \,8x\,.\cos \,16x$$ then $$f'\left( {\frac{\pi }{4}} \right)$$ is :
A.
$$\sqrt 2 $$
B.
$$\frac{1}{{\sqrt 2 }}$$
C.
1
D.
none of these
Answer :
$$\sqrt 2 $$
Solution :
$$\eqalign{
& f\left( x \right) = \frac{{2\sin \,x.\cos \,x.\cos \,2x.\cos \,4x.\cos \,8x.\cos \,16x}}{{2\sin \,x}} = \frac{{\sin \,32x}}{{{2^5}\sin \,x}} \cr
& \therefore f'\left( x \right) = \frac{1}{{32}}.\frac{{32\cos \,32x.\sin \,x - \cos \,x.\sin \,32x}}{{{{\sin }^2}x}} \cr
& \therefore f'\left( {\frac{\pi }{4}} \right) = \frac{{32.\frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}.0}}{{32.{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}} \cr} $$