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