What is the derivative of #sqrtx^2#?

2 Answers
Jun 22, 2015

The root of a square is the number itself.

Explanation:

#sqrtx^2=x#

The derivative of this is #1#

Jun 22, 2015

If #f(x) = sqrtx^2#, the #f(x) = x#, so #f'(x) = 1#, but

Explanation:

Be careful:

#g(x)=sqrt(x^2) = abs(x) = {(x, "if",x >= 0),(-x,"if",x<0) :}#

So #g'(x) = {(1, "if",x > 0),(-1,"if",x<0) :}#

#g'(0)# does not exist.