Zadatak 0038

Test: MPsIVKVI_008

 

Ako je \(f(x)=\frac{\tan x}{1-\tan x}\) naći \(f'(\frac{\pi }{6})\).

 

Rešenje:


 

 

\(f'(x)=\frac{\frac{1}{\cos ^2x}\cdot (1-\tan x)-(-\frac{1}{\cos ^2x})\cdot \tan x}{(1-\tan x)^2}\) \(=\frac{1}{\cos ^2x}\cdot \frac{1-\tan x+\tan x}{\left ( \frac{\cos x-\sin x}{\cos x} \right )^2}\) \(=\frac{1}{\cos ^2x-2\cos x\sin x+\sin ^2x}=\frac{1}{1-\sin 2x};\) 

 

\(f'(\frac{\pi }{6})=\frac{1}{1-\frac{\sqrt{3}}{2}}\) \(=\frac{2}{2-\sqrt{3}}\cdot \frac{2+\sqrt{3}}{2+\sqrt{3}}=4+2\sqrt{3}\).