If `f(x) = cos x cos 2x cos 4x cos (8x). cos 16x` then find `f' (pi/4)`

Options

A. `sqrt(2)`
B. `(1)/(sqrt(2))`
C. 0
D. `(sqrt(3))/(2)`

Answer

C

Explanation

Clearly , ` f((pi)/(4)) = cos""(po)/(4) cos "" (pi)/(2) cos pi cos 2 pi cos 4 pi = 0 `
On taking log on given function , we get
` log f(x) = log cos x + log cos 2x + log cos 4x + log cos 8x + log cos 16 x `
On differentiating both sides w.r.t.'x', we get
`rArr (1)/(f(x)) = - [tan x + 2 tan 2x + 4 tan 4x - 8 tan 8x + 16 tan 16 x `
`rArr f'(x) = - f(x)[tanx + 2 tan 2x + 4 tan 4x + 8 x + 16 tan 16 x `
`rArr f'((pi)/(4)) = - f((pi)/(4))`
`[tan""(pi)/(4) + 2 tan ""(pi)/(2) + 4 tan pi + 16 tan 4 pi ] = 0 `
` [ therefore ((pi)/(4))= 0 ]`