Coi như chỉ xét tại những điểm hàm có đạo hàm (tức \(x\ne\pm1\))
\(\left(\left|1+x\right|\right)'=\left(\sqrt{\left(1+x\right)^2}\right)'=\dfrac{1+x}{\left|1+x\right|}\) ; \(\left(\left|1-x\right|\right)'=\left(\sqrt{\left(1-x\right)^2}\right)'=-\dfrac{1-x}{\left|1-x\right|}\)
Do đó:
\(f'\left(x\right)=\dfrac{\left(\dfrac{1+x}{\left|1+x\right|}+\dfrac{1-x}{\left|1-x\right|}\right)\left(\left|1+x\right|+\left|1-x\right|\right)-\left(\left|1+x\right|-\left|1-x\right|\right)\left(\dfrac{1+x}{\left|1+x\right|}-\dfrac{1-x}{\left|1-x\right|}\right)}{\left(\left|1+x\right|+\left|1-x\right|\right)^2}\)
\(=\dfrac{\dfrac{2\left(1+x\right)\left|1-x\right|}{\left|1+x\right|}+\dfrac{2\left(1-x\right)\left|1+x\right|}{\left|1-x\right|}}{\left(\left|1+x\right|+\left|1-x\right|\right)^2}\)
\(=\dfrac{4\left(1-x^2\right)}{\left|1-x^2\right|\left(\left|1+x\right|+\left|1-x\right|\right)^2}\)
