Cho hàm số \(y=\varphi\left(t\right)=\sqrt{1+\cos^2t^2}\) . Tính \(\varphi'\left(\dfrac{\sqrt{\pi}}{2}\right)\).
\(-\dfrac{\sqrt{6\pi}}{3}\).\(-\dfrac{\sqrt{6\pi}}{6}\).\(-\dfrac{\sqrt{3\pi}}{2}\).\(-\dfrac{\sqrt{3\pi}}{3}\).Hướng dẫn giải:Đặt \(u=1+\cos^2t^2=1+\dfrac{1+\cos2t^2}{2}=\dfrac{3}{2}+\dfrac{1}{2}\cos2t^2\) thì \(u'=\dfrac{1}{2}.\left(-\sin2t^2\right).\left(4t\right)=-2t.\sin2t^2\)
\(\varphi\left(t\right)=\sqrt{u},\varphi'\left(t\right)=\dfrac{u'}{2\sqrt{u}}=\dfrac{-2t.\sin2t^2}{2\sqrt{1+\cos^2t^2}}=-\dfrac{t.\sin2t^2}{\sqrt{1+\cos^2t^2}};\varphi'\left(\dfrac{\sqrt{\pi}}{2}\right)=-\dfrac{\dfrac{\sqrt{\pi}}{2}\sin\dfrac{\pi}{2}}{\sqrt{1+\cos^2\dfrac{\pi}{4}}}=-\dfrac{\sqrt{\pi}}{\sqrt{6}}=-\dfrac{\sqrt{6\pi}}{6}\)
