Cho hàm số \(y=\varphi\left(x\right)=\sqrt{x^2+1}-\ln\left(\frac{1+\sqrt{x^2+1}}{x}\right)\). Tính \(\varphi'\left(2\right)\) .
\(\frac{\sqrt{3}}{2}\) \(1\) \(\frac{\sqrt{5}}{2}\) \(\frac{\sqrt{6}}{2}\) Hướng dẫn giải:Ta có \(\left(1+\sqrt{x^2+1}\right)'=\left(\sqrt{x^2+1}\right)'=\dfrac{x}{\sqrt{x^2+1}},\)
Vì \(\varphi\left(x\right)=\sqrt{x^2+1}-\ln\left(1+\sqrt{x^2+1}\right)+\ln x\Rightarrow\varphi'\left(x\right)=\dfrac{x}{\sqrt{x^2+1}}-\dfrac{x}{\sqrt{x^2+1}}:\left(1+\sqrt{x^2+1}\right)+\dfrac{1}{x}\)
Do đó \(\varphi'\left(2\right)=\dfrac{2}{\sqrt{5}}-\dfrac{2}{\sqrt{5}\left(1+\sqrt{5}\right)}+\dfrac{1}{2}=\dfrac{2}{\sqrt{5}}\left(1-\dfrac{1}{1+\sqrt{5}}\right)+\dfrac{1}{2}=\dfrac{2}{1+\sqrt{5}}+\dfrac{1}{2}=\dfrac{5+\sqrt{5}}{2\left(1+\sqrt{5}\right)}=\dfrac{\sqrt{5}}{2}\)