Answer:
No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R
Step-by-step explanation:
Assuming: the function is
in [0,1]
And rewriting it for the sake of clarity:
Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer
1) A function is considered to be differentiable if, and only if both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]
The limit to the left


g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)
3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R
Because this is the same as to calculate the limit from the left and right side, of g(x).
![f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]](https://tex.z-dn.net/?f=f%27%28c%29%3D%5Clim_%7Bx%5Crightarrow%20c%7D%5Cleft%20%5B%5Cfrac%7Bf%28b%29-f%28a%29%7D%7Bb-a%7D%20%5Cright%20%5D%5C%5C%5C%5Cg%27%280%29%3D%5Clim_%7Bx%5Crightarrow%200%7D%5Cleft%20%5B%5Cfrac%7Bg%28b%29-g%28a%29%7D%7Bb-a%7D%20%5Cright%20%5D%5C%5C%5C%5Cg%27%281%29%3D%5Clim_%7Bx%5Crightarrow%201%7D%5Cleft%20%5B%5Cfrac%7Bg%28b%29-g%28a%29%7D%7Bb-a%7D%20%5Cright%20%5D)
This is what the Bilateral Theorem says:
