What is the derivative of #sqrtx^2#?

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Answer 1 · 2015-06-22T14:13:10

The root of a square is the number itself.

#sqrtx^2=x#

The derivative of this is #1#

Answer 2 · 2015-06-22T14:39:15

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

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.