Wavefunction and its derivative vanish at same point It is stated that if $\psi(x_0)=\psi'(x_0)=0$ at some $x_0$, then $\psi(x)=0,\forall x$.
Hence, we cannot have the wavefunction and its derivative vanishing at the same point.
My question is, how to prove the above statement? 
 A: 
It is stated that if $\psi(x_0)=\psi'(x_0)=0$ at some $x_0$, then $\psi(x)=0, \ \forall x$. Hence, we cannot have the wavefunction and its derivative vanishing at the same point.

This is not true, but I believe I know what you mean.

In 1D, the time-independent Schrodinger equation is a linear, 2nd order ODE of the form
$$-\psi'' + V(x) \psi = E\psi$$
or
$$\psi'' + (E-V)\psi = 0$$
The existence and uniqueness theorem for a linear, 2nd order ODE is usually stated as follows:

Let $p(x), q(x),$ and $g(x)$ be continuous functions on some open interval $I$.  Given any $x_0\in I$, there is a unique function $y(x)$ defined on $I$ which satisfies the differential equation
  $$y''(x) + p(x) y'(x) + q(x) y(x) = g(x)$$
  and the initial conditions $y(x_0)=y_0$, $y'(x_0)=y'_0$.

The proof of this is fairly involved.  If you want to understand it, Coddington's book on ODEs is pretty inexpensive and discusses it pretty thoroughly (Chapter 6, Sections 7 and 8).
If we take that at face value, then the result you're looking for is immediate.  Note that if $y_0=y'_0=0$, then the trivial function $y(x)=0$ satisfies both the Schrodinger equation and the initial conditions; therefore, it follows from the uniqueness theorem that it is the only function which does so, and must therefore be the solution we're looking for.

That being said, the reason that the original claim is not true is simply that in general, a wave function need not satisfy the time-independent Schrodinger equation.  The TISE is only satisfied by energy eigenfunctions.
