what is the condition for this statement to be true?
Use the definition of the derivative.
$\displaystyle f'(x)=\lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$
Multilpy the numerator and denominator by the conjugate to get
$\displaystyle f'(x)=\lim_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\left( \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\ri ght)=\lim_{h \to 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}=\frac{1}{2\sqrt{x}}$
Good luck
I think that this works as long as $\displaystyle x \ne 0$.
Here is why: if It is differentiable at zero then this limit must exist
$\displaystyle f'(0)=\lim_{h \to 0^+}\frac{\sqrt{0+h}-\sqrt{0}}{h}=\lim_{h \to 0^+}\frac{\sqrt{h}}{h}=\lim_{h \to 0^+}\frac{1}{\sqrt{h}}=\infty$
So $\displaystyle f(x)=\sqrt{x}$ is only differentable on $\displaystyle (0,\infty)$
Edit: I misread your other thread.
Okay so I just saw your other thread.
I think you want to show that
$\displaystyle \int_{0}^{1}\sqrt{x}dx$ is Riemann integrabole on [0,1]
Here is one method let $\displaystyle \epsilon > 0$
$\displaystyle f(x)=\sqrt{x}$ is continuous on $\displaystyle [0,1]$ so it is uniformly continuous on $\displaystyle [0,1]$
Since f is unifomly continous on [0,1] there exists a $\displaystyle \delta$ such that $\displaystyle |f(x)-f(y)|< \epsilon$
Now let $\displaystyle P=\{x_0,x_1,...x_n\} $be any partition of [0,1] such that $\displaystyle |x_{j}-x_{j-1}|<\delta$. Then by the extreme value theorem there are points $\displaystyle x_M,x_m$ are the max and min on $\displaystyle x_j,x_{j-1}$ for each j.
Since the mesh of P is less than delta we have $\displaystyle |x_M-x_m|< \delta$ so $\displaystyle |f(x_M)-f(x_m) |< \epsilon$
So the upper and lower sums are
$\displaystyle U(f,P)=\sum_{j=1}^{n}f(x_M)(x_j-x_{j-1})$
$\displaystyle L(f,P)=\sum_{j=1}^{n}f(x_m)(x_j-x_{j-1})$
$\displaystyle U(f,P)-L(f,P)=\sum_{j=1}^{n}(f(x_M)-f(x_m)(x_j-x_{j-1})< \epsilon \sum_{j=1}^{n}(x_j-x_{j-1})=\epsilon(1-0)=\epsilon$