Cho hàm số y = f(x) = căn bậc hai tan x + cot x . Tính f ' ( pi/4)

\({f^\prime }(x) = \frac{{1 + {{\tan }^2}x - \left( {1 + {{\cot }^2}x} \right)}}{{2\sqrt {\tan x + \cot x} }} = \frac{{{{\tan }^2}x - {{\cot }^2}x}}{{2\sqrt {\tan x + \cot x} }} \Rightarrow {f^\prime }(0) = \frac{{{{\tan }^2}\frac{\pi }{4} - {{\cot }^2}\frac{\pi }{4}}}{{2\sqrt {\tan \frac{\pi }{4} + \cot \frac{\pi }{4}} }} = 0\)