Cho hàm số \(y=f\left(x\right)=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\ln\left(x+\sqrt{x^2-a^2}\right)\). Tính \(f'\left(2a\right)\).
\(a\sqrt{2}\) \(a\) \(a\sqrt{3}\) \(2a\) Hướng dẫn giải:Ta có \(\left(\sqrt{x^2-a^2}\right)'=\dfrac{x}{\sqrt{x^2-a^2}}\), \(\left(x+\sqrt{x^2-a^2}\right)'=1+\dfrac{x}{\sqrt{x^2-a^2}}=\dfrac{x+\sqrt{x^2-a^2}}{\sqrt{x^2-a^2}}\).
Do đó \(\left(\dfrac{x}{2}\sqrt{x^2-a^2}\right)'=\dfrac{1}{2}\sqrt{x^2-a^2}+\dfrac{x}{2}.\dfrac{x}{\sqrt{x^2-a^2}}=\dfrac{2x^2-a^2}{2\sqrt{x^2-a^2}}\) và \(\left(\ln\left(x+\sqrt{x^2-a^2}\right)\right)'=\dfrac{x+\sqrt{x^2-a^2}}{\sqrt{x^2-a^2}}:\left(x+\sqrt{x^2-a^2}\right)=\dfrac{1}{\sqrt{x^2-a^2}}\)
\(y'=\dfrac{2x^2-a^2}{2\sqrt{x^2-a^2}}-\dfrac{a^2}{2\sqrt{x^2-a^2}}=\sqrt{x^2-a^2}\). Vì vậy \(f'\left(2a\right)=\sqrt{4a^2-a^2}=a\sqrt{3}\)
