1/|x| đạo hàm bằng j ạ ?

\(\left({\dfrac{1}{|x|}}\right)'=\dfrac{(1)'|x|-1.|x|'}{|x|^2}=\dfrac{-|x|'}{x^2}\)

Nếu \(x>0\) thì \(\) \(\left({\dfrac{1}{|x|}}\right)'=\dfrac{-(x)'}{x^2}=\dfrac{-1}{x^2}\).

Nếu \(x<0\) thì \(\) \(\left({\dfrac{1}{|x|}}\right)'=\dfrac{-(-x)'}{x^2}=\dfrac{1}{x^2}\).

\[\begin{array}{l} y = \frac{1}{{\left| x \right|}} = \left\{ \begin{array}{l} \frac{1}{x}\,\,\,\,khi\,\,\,x > 0\\ - \frac{1}{x}\,\,\,\,khi\,\,\,x < 0 \end{array} \right.\\ \Rightarrow y' = \left\{ \begin{array}{l} - \frac{1}{{{x^2}}}\,\,\,khi\,\,\,x > 0\\ \frac{1}{{{x^2}}}\,\,\,khi\,\,\,x < 0 \end{array} \right.. \end{array}\]