Giải thích các bước giải:
Ta có:
$y=\dfrac{x}{\sqrt{x^2+1}}$
$\to y'=(\dfrac{x}{\sqrt{x^2+1}})'$
$\to y'=\dfrac{x'\sqrt{x^2+1}-\left(\sqrt{x^2+1}\right)'\:x}{\left(\sqrt{x^2+1}\right)^2}$
$\to y'=\dfrac{1\cdot \sqrt{x^2+1}-\dfrac{x}{\sqrt{x^2+1}}x}{\left(\sqrt{x^2+1}\right)^2}$
$\to y'=\dfrac{1}{\left(x^2+1\right)\sqrt{x^2+1}}$
$\to y'(2\sqrt{2})=\dfrac{1}{\left((2\sqrt{2})^2+1\right)\sqrt{(2\sqrt{2})^2+1}}=\dfrac1{27}$