Cho hàm số \(f\left(x\right)=-\dfrac{\cos x}{2\sin^2x}+\ln\sqrt{\dfrac{1+\cos x}{\sin x}}\). Tính \(f'\left(\frac{\pi}{4}\right)\) .
1\(\frac{1}{2}\)\(\sqrt{2}\)\(\frac{1}{\sqrt{2}}\)Hướng dẫn giải:- Tính \(\left(\dfrac{\cos x}{\sin^2x}\right)'\) : đặt \(u=\cos x,v=\sin^2x\) thì \(u'=-\sin x,v'=2\sin x\cos x\) và \(u'v-uv'=-\sin^3x-2\sin x\cos^2x\). Do đó
\(\left(\dfrac{\cos x}{\sin^2x}\right)'=\dfrac{-\sin^3x-2\sin x\cos^2x}{\sin^4x}=-\dfrac{\sin^2x+2\cos^2x}{\sin^3x}=-\dfrac{1+\cos^2x}{\sin^3x}\)
- Tính \(\left(\ln\sqrt{\dfrac{1+\cos x}{\sin x}}\right)'\):
Ta có \(\dfrac{1+\cos x}{\sin x}=\dfrac{2\cos^2\dfrac{x}{2}}{2\sin\dfrac{x}{2}\cos\dfrac{x}{2}}=\cot\dfrac{x}{2}\) nên \(\ln\sqrt{\dfrac{1+\cos x}{\sin x}}=\dfrac{1}{2}\ln\dfrac{1+\cos x}{\sin x}=\dfrac{1}{2}\ln\cot\dfrac{x}{2}\), do đó
\(\left(\ln\sqrt{\dfrac{1+\cos x}{\sin x}}\right)'=\dfrac{1}{2}\left(\ln\cot\dfrac{x}{2}\right)'=\dfrac{1}{2}.\dfrac{-\dfrac{1}{2}.\dfrac{1}{\sin^2\dfrac{x}{2}}}{\cot\dfrac{x}{2}}=\dfrac{-1}{4\sin^2\dfrac{x}{2}\cot\dfrac{x}{2}}=\dfrac{-1}{4\sin\dfrac{x}{2}\cos\dfrac{x}{2}}=\dfrac{-1}{2\sin x}\)
Vì vậy \(f'\left(x\right)=\dfrac{1+\cos^2x}{2\sin^3x}-\dfrac{1}{2\sin x}=\dfrac{1+\cos^2x-\sin^2x}{2\sin^3x}=\dfrac{\cos^2x}{\sin^3x}=\dfrac{\cot^2x}{\sin x}\).
Khi \(x=\dfrac{\pi}{4}\) thì \(\cot x=1,\sin x=\dfrac{1}{\sqrt{2}}\) do đó \(f'\left(\dfrac{\pi}{4}\right)=\sqrt{2}\).