If $f(0)=0$, can I say that $f'(0)=0$ as well? Just a simple question on derivatives:
If $f(0)=0$, then can I say that $f'(0)=0$ or it is completely wrong?
But, if $f(x)=0$, then can I say that $f'(x)=0$ or and this is completely wrong?
 A: Hint:
One counterexample would be $f(x) =\sin x$.
Note that $f(0)=0$, but $f’(0) =\cos x\lvert_{x=0} = 1$.
A: 1) consider f(x)=sinx, f(0)=0 however f'(x)=cosx so f'(0)=1
2)if f(x)=0, then for all x f(x)=0 so the function doesn't vary with x, so the rate of change of the function is 0 i.e f'(x)=0
A: A function property holding at a single point or at all points are two very different issues.
The derivative of a function requires the knowledge of infinitely many values within a neighborhhood of the point (it takes the computation of a limit).
So in the first case, the values of $f(0)$ and $f'(0)$ are completely independent (because you only specify one value of $f$), while in the second, we obviously have $0'=0$ (you specified all values of $f$).

Additional material:
A point where $f(x)=0$ is a root of the function, and one where $f'(x)=0$ can correspond to a local extremum. There is no connection between the locations of the roots and those of the extrema.
The function $f(x)=0$ is one of the most "dull".
A: Note that $g(x)=f(x)+c$ has the same derivative as $f(x)$. 
So I can take any function I like, with $f'(a)$ what I like, and then construct $g(x)=f(x)-f(a)$ which has $g'(a)=f'(a)$ the number I originally thought of, and $g(a)=f(a)-f(a)=0$
A: A generalization: you can even construct a function with $f(0)$, $f^{(1)}(0)$,..., $f^{(N)}(0)$ of your choice through
$$f(x)=\sum_{n=0}^{N}\frac{f^{(n)}(0)}{n!}x^n$$
