Question

Sin4x - cosx ---------------- = f(x) f^1(π/4) what is the derivative? tanx

Answer 1

I think you are asked to find the value of the first derivative of f(x) at π/4. Given

$f(x) = \dfrac{\sin(4x)-\cos(x)}{\tan(x)}$

use the quotient to differentiate and you get

$f'(x) = \dfrac{\tan(x)(4\cos(4x)+\sin(x))-(\sin(4x)-\cos(x))\sec^2(x)}{\tan^2(x)}$

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