I think you are asked to find the value of the first derivative of f(x) at π/4. Given
[tex]f(x) = \dfrac{\sin(4x)-\cos(x)}{\tan(x)}[/tex]
use the quotient to differentiate and you get
[tex]f'(x) = \dfrac{\tan(x)(4\cos(4x)+\sin(x))-(\sin(4x)-\cos(x))\sec^2(x)}{\tan^2(x)}[/tex]
Then at x = π/4, you have
tan(π/4) = 1
cos(4•π/4) = cos(π) = -1
sin(π/4) = 1/√2
sin(4•π/4) = sin(π) = 0
cos(π/4) = 1/√2
sec(π/4) = √2
==> f ' (π/4) = (1•(-4 + 1/√2) - (0 - 1/√2)•(√2)²) / 1² = -4 + 1/√2 + √2
