f(x)=I x^2-9 I can be written as
f(x) = [ ( x^2 - 9)^2 ]^(1/2)
So....using the Chain Rule twice, we have :
f ' (x) = (1/2) [ (x^2 - 9)^2] ^(-1/2) [ 2 (x^2 - 9) *2x] =
2x [ x^2 - 9] / [(x^2 - 9)^2] ^(1/2) =
2x[x^2 - 9] / l x^2 - 9 l
Note that if x = -3 or if x = 3, the derivative is undefined, because we can't divide by 0
Notice the "sharp points" at x = -3 and x = 3 on the graph : https://www.desmos.com/calculator/ee9uapzsmi
The derivative is undefined at these points.......so.......this funnction is differentiable at all x except at x = -3 and x = 3
