Cho hàm số \(f\left(x\right)=\dfrac{\cos^2x}{1+\sin^2x}\)
Tính giá trị biểu thức \(f\left(\dfrac{\pi}{4}\right)-3f'\left(\dfrac{\pi}{4}\right)\).
\(-3\).\(\dfrac{8}{3}\).\(3\).\(-\dfrac{8}{3}\).Hướng dẫn giải:Ta có \(f\left(x\right)=\dfrac{\cos^2x}{1+\sin^2x}=\dfrac{1}{\dfrac{1}{\cos^2x}+\tan^2x}=\dfrac{1}{2\tan^2x+1}\). Đặt \(u=2\tan^2x+1\) thì \(f\left(x\right)=\dfrac{1}{u},f'\left(x\right)=-\dfrac{u'}{u^2}\) .
Tính \(u'\): \(u'=4\tan x\left(\tan x\right)'=4\tan x\left(1+\tan^2x\right)\). Vì vậy \(f'\left(x\right)=-\dfrac{4\tan x\left(1+\tan^2x\right)}{\left(2\tan^2x+1\right)^2}\).
Khi \(x=\dfrac{\pi}{4}\) thì \(\tan x=1\) nên \(f\left(\dfrac{\pi}{4}\right)-3f'\left(\dfrac{\pi}{4}\right)=\dfrac{1}{2+1}-3\left(-\dfrac{4.1\left(1+1\right)}{\left(2+1\right)^2}\right)=\dfrac{1}{3}+\dfrac{8}{3}=3\)
Đáp số:\(3\)