Cho hàm số \(y=f\left(t\right)=\ln\left(\frac{2+\tan t}{2-\tan t}\right)\). Trong các khẳng định sau, khẳng định nào sai?
\(f'\left(\frac{\pi}{3}\right)=16\) \(f'\left(\frac{\pi}{6}\right)=\frac{1}{2}\) \(f'\left(\frac{\pi}{4}\right)=\frac{8}{3}\) \(f'\left(0\right)=1\) Hướng dẫn giải:Đặt \(u=\dfrac{2+\tan x}{2-\tan x}=\dfrac{4-\left(2-\tan x\right)}{2-\tan x}=4.\dfrac{1}{2-\tan x}-1\Rightarrow u'=-4.\dfrac{\left(2-\tan x\right)'}{\left(2-\tan x\right)^2}=\dfrac{4\left(1+\tan^2x\right)}{\left(2-\tan x\right)^2}\)
\(f\left(x\right)=\ln u,f'\left(x\right)=\dfrac{u'}{u}=\dfrac{4\left(1+\tan^2x\right)}{\left(2-\tan x\right)^2}.\dfrac{2-\tan x}{2+\tan x}=\dfrac{4\left(1+\tan^2x\right)}{4-\tan^2x}\)
\(f'\left(\dfrac{\pi}{3}\right)=\dfrac{4\left(1+3\right)}{4-3}=16,f'\left(\dfrac{\pi}{6}\right)=\dfrac{4\left(1+\dfrac{1}{3}\right)}{4-\dfrac{1}{3}}=\dfrac{16}{11},f'\left(\dfrac{\pi}{4}\right)=\dfrac{4\left(1+1\right)}{4-1}=\dfrac{8}{3},f'\left(0\right)=\dfrac{4\left(1+0\right)}{4-0}=1\)