Đạo hàm của hàm số: f(x) = (x^2 + 1)/(căn bậc hai (a^2 + x^2))

* Hướng dẫn giải

Chọn B

$\begin{array}{l}y\text{'}=\frac{2x.\sqrt{a{}^2+\text{}{x}^2}+(x^2+\text{}1).\frac{1}{2\sqrt{a^2+\text{}{x}^2}}.(a^2+\text{}{x}^2)\text{'}}{a^2+\text{\hspace{0.17em}\hspace{0.17em}}{x}^2}\\ =\frac{2x.\sqrt{a{}^2+\text{}{x}^2}+\frac{x(\text{}{x}^2+1)}{\sqrt{a^2+\text{}{x}^2}}}{a^2+\text{\hspace{0.17em}\hspace{0.17em}}{x}^2}=\frac{2x(a^2+\text{}{x}^2)+\text{}x(x^2+1)\text{}}{(a^2+x^2).\sqrt{a^2+\text{}{x}^2}}\\ =\frac{x(3x^2+\text{}2a^2+1)\text{}}{(a^2+x^2).\sqrt{a^2+\text{}{x}^2}}\end{array}$