If $f(x)=\int_{0}^{x}\sqrt{f(t)}\,dt$ then $f(6)$ is? 
let $f:[0,\infty) \rightarrow[0,\infty]$ be continuous on $[0,\infty]$
  and differentiable on $(0,\infty)$. if $f(x)=\int_{0}^{x}\sqrt{f(t)}dt$
  then $f(6)$ is ?

$f(0)=0$
$f'(x)=\sqrt{f(x)}\implies f'(6)=\sqrt{f(6)}\implies f'(6)^2=f(6)$
Now I am trying to find out $f'(6) $ somehow
$f'(6)=\lim_{x \rightarrow 6}\dfrac{f(x)-f(6)}{x-6}=\lim_{x \rightarrow 6}\dfrac{f(x)-f'(6)^2}{x-6}$
$f'(0)=\lim_{x \rightarrow 0}\dfrac{f(x)-f(0)}{x-0}$
I am missing something and I am not able to solve this problem by above steps.
 Any hint? 
 A: Suppose $f(x) = 0$ for some $x \in (0,\infty)$. Then, $f$ is the integral of a non-negative function from $0$ to $x$, so this forces $f(y) = 0$ for all $y \leq x$.
On the contrary, $f$ being the integral of a non-negative function is increasing. Therefore, if $x^* = \inf\{y : f(y) = 0\}$, then $f$ is positive beyond $x^*$ and zero before (and including) that point.

KEY POINT : Once $f$ is shown to be positive, so is the square root, and therefore the derivative exists since we may divide by $\sqrt f$.
Since $f' = \sqrt f$, $f'$ is differentiable in $(x^*,\infty)$ (and positive on this interval, since we are considering the positive square root), with $f'' = (\sqrt f)' = \frac{1}{2\sqrt f} \times f' = \frac 12$ everywhere on $(x^*,\infty)$. Note that $f''(x^*) = 0$, from the fact that $f'$ is zero on $[0,x^*]$. 
Thus, $f''$ is constant on $(x^*,\infty)$ and equal to $\frac 12$.We may solve this equation and obtain $f''$ as some quadratic on this interval, since $f '' = \frac 12 \implies f = \frac{x^2}{4} + c_1x+c_2$, with the conditions at $x^*$. You can find what $c_1,c_2$ are.
Thus, if $f$ is as above, then we may characterize $f$ as being $0$ up till some $x^*$ and then the quadratic we obtained above. In particular, $f(6)$ does not take one svalue, but a range of values depending on which $f$ we are referring to. I am sure you will be able to deduce the exact range of values it lies in from here.

For completeness, I add the solution : For any $x^*$, the function $f_{x^*}(x) = \mathbf 1_{x > x^*} \frac{(x-x^*)^2}{4}$ satisfies the given conditions. Conversely, any function satisfying the given conditions must be of this form.
From here, one sees that $f(6)$ may take any of the values $\mathbf 1_{6 > x^*} \frac{(6-x^*)^2}{4}$, from where the decreasing nature of this function in $x^*$ gives the answer as $f(6) \in [0,9]$. Note that if $x^* = 0$ then $f(6) = 9$,  which is the "standard" answer under the common mistake of not realizing that $f$ needs to be positive for $\sqrt f$ to be invertible.
A: Assume $f(x) > 0$.  Upon differentiation, we have $y' = \sqrt{y} $ where $y=f(x)$. We have $\int \frac{dy }{y^{1/2}} = x + C $. Thus $2 \sqrt{y} = x + C $ And $C=0$ as $f(0)=0$. Thus, $y = \frac{x^2}{4}$. It follows that $f(6) = 9$
Added: See Clement C's comments below for a more rigorous solutions when $f(x) > 0$ is not assumed.
A: An alternative method (the case $f=0$ is trivial; see @ClementC’s comment for slight variations of this answer where $f>0$ isn’t assumed for $x>0$):
You already know $f’(x)=\sqrt{f(x)}$. Since the composition of differentials functions is differentiable (this is where we use the fact that $f(x)>0$ for $x>0$), we deduce that $f’’(x)=\frac12$. Integrating once, we get $f’(x)=\frac{x}{2}+C$, and using the fact that $f(0)=0$, we can see that $f’(0)=0$, meaning $C=0$. Integrating once more and using the initial value, we have $$f(x)=\frac{x^2}{4}.$$
