Cho hàm số \(y = f\left( x \right) = \sqrt {{x^2}...

C

- Hướng dẫn giải

\(\begin{array}{l}
y' = \frac{1}{{2\sqrt {{x^2} + 1} }}.\left( {{x^2} + 1} \right)' = \frac{{2x}}{{2\sqrt {{x^2} + 1} }} = \frac{x}{{\sqrt {{x^2} + 1} }}\\
y'' = \frac{{\sqrt {{x^2} + 1}  - x.\frac{{\left( {{x^2} + 1} \right)'}}{{2\sqrt {{x^2} + 1} }}}}{{{x^2} + 1}} = \frac{{\sqrt {{x^2} + 1}  - \frac{{{x^2}}}{{\sqrt {{x^2} + 1} }}}}{{{x^2} + 1}}\\
 = \frac{1}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }}\\
 \Rightarrow y.y' = x,{y^2}.y'' = \frac{1}{{\sqrt {{x^2} + 1} }}
\end{array}\)