Cho hàm số \(f\left(x\right)=\dfrac{1}{\sqrt{3}}\ln\left(\dfrac{\tan\dfrac{x}{2}+2-\sqrt{3}}{\tan\dfrac{x}{2}+2+\sqrt{3}}\right)\)
Tính \(f'\left(\frac{\pi}{2}\right)\).
\(\frac{1}{3}\) \(\frac{2}{3}\) \(1\) \(\frac{4}{3}\) Hướng dẫn giải:Cách 1: Dùng MTCT Đáp số đúng là \(\frac{1}{3}.\)
Cách 2: Đặt \(u=\dfrac{\tan\dfrac{x}{2}+2-\sqrt{2}}{\tan\dfrac{x}{2}+2+\sqrt{3}}\) thì \(f\left(x\right)=\dfrac{1}{\sqrt{3}}\ln u\) nên \(f'\left(x\right)=\dfrac{1}{\sqrt{3}}.\dfrac{u'}{u}\).
Tính \(u'\) : Đặt \(t=\tan\dfrac{x}{2}\) thì \(u=\dfrac{t+2-\sqrt{3}}{t+2+\sqrt{3}}\) và \(u'=\left(\dfrac{t+2-\sqrt{3}}{t+2+\sqrt{3}}\right)'.t'\).
Vì \(t=\tan\dfrac{x}{2}\Rightarrow t'=\dfrac{1}{2}\left(1+\tan^2\dfrac{x}{2}\right)\) và vì \(\left(\dfrac{t+2-\sqrt{3}}{t+2+\sqrt{3}}\right)'=\dfrac{2\sqrt{3}}{\left(t+2+\sqrt{3}\right)^2}\) nên \(u'=\dfrac{2\sqrt{3}.\dfrac{1}{2}\left(1+\tan^2\dfrac{x}{2}\right)}{\left(\tan\dfrac{x}{2}+2+\sqrt{3}\right)^2}=\dfrac{\sqrt{3}\left(1+\tan^2\dfrac{x}{2}\right)}{\left(\tan\dfrac{x}{2}+2+\sqrt{3}\right)^2}\)
Do đó
\(f'\left(x\right)=\dfrac{1}{\sqrt{3}}.\dfrac{\sqrt{3}\left(1+\tan^2\dfrac{x}{2}\right)}{\left(\tan\dfrac{x}{2}+2+\sqrt{3}\right)^2}.\dfrac{\tan\dfrac{x}{2}+2+\sqrt{3}}{\tan\dfrac{x}{2}+2-\sqrt{3}}\)\(=\dfrac{1+\tan^2\dfrac{x}{2}}{\left(\tan\dfrac{x}{2}+2\right)^2-3}\)
Chú ý rằng khi \(x=\dfrac{\pi}{2}\) thì \(\tan\dfrac{x}{2}=\tan\dfrac{\pi}{4}=1\) nên \(f'\left(\dfrac{\pi}{2}\right)=\dfrac{1+1^2}{\left(1+2\right)^2-3}=\dfrac{1}{3}\)
Đáp số: \(\dfrac{1}{3}\)