फलन `f(x)=x।x।` के लिए बिंदु x = 0 पर अवकलनीयता पर विचार कीजिए।

Explanation

यहाँ `f(x)=x|x|={(x^(2)",",x ge0),(-x^(2)",",x lt 0):} " " [ because |x|={(x",",x ge0),(-x",",x lt 0):}]`
` f(0)=0^(2)=0`
` Rf'(0)=lim_(h to 0) (f(0+h)-f(0))/(h)`
`=lim_ ( h to 0) ((0+h)^(2)-0)/(h), [ because x =0+h gt 0]`
`=lim_(h to 0) (h^(2))/(h)`
`=lim_(h to 0) h=0.`
और `L f'(0)=lim_(h to 0) (f(0-h)-f(0))/(-h)`
`=lim_(h to 0) (-(0-h)^(2)-0)/(-h), [ because x = 0-h lt 0]`
`=lim_(h to 0) (-h^(2))/(h)`
`=lim_(h to 0)-h=0.`
`therefore R f'(0)=L f'(0).`
अतः f(x) बिंदु x = 0 पर अवकलनीय है।