Giải thích các bước giải:
Ta có:
$y=\sqrt{x+\sqrt{x^2+1}}$
$\to y'=(\sqrt{x+\sqrt{x^2+1}})'$
$\to y'=\dfrac{1}{2\sqrt{x+\sqrt{x^2+1}}}\left(x+\sqrt{x^2+1}\right)'\:$
$\to y'=\dfrac{1}{2\sqrt{x+\sqrt{x^2+1}}}\left(1+\dfrac{x}{\sqrt{x^2+1}}\right)$
$\to y'=\dfrac{\sqrt{\sqrt{x^2+1}+x}}{2\sqrt{x^2+1}}$
$\to y'(\sqrt3)=\dfrac{\sqrt{\sqrt{(\sqrt3)^2+1}+(\sqrt3)}}{2\sqrt{(\sqrt3)^2+1}}$
$\to y'(\sqrt3)=\dfrac{\sqrt{2+\sqrt{3}}}{4}$