Derivation of Nyquist Frequency and Sampling Theorem I have been looking through different sites and questions over the internet about Sampling theory, but couldn’t find the clear definition of how nyquist frequency condition is derived? It would be great if someone could direct me to the derivation of the condition that sampling frequency should >= to maximum frequency in the signal
It really boils down to aliasing.  In continuous-time, if you have any two signals $x_1(t) = \sin(2 \pi F_1 t)$ and $x_2(t) = \sin(2 \pi F_2 t)$, then as long as $F_1$ and $F_2$ are distinct, the signals are, too.

But consider sampling at some time interval $T_s$, so that the sampled signals are $x_1(k) = \sin(2 \pi F_1 T_s k)$ and $x_2(k) = \sin(2 \pi F_2 T_s k)$.  If you have a condition where $2 \pi F_1 = 2 \pi F_2 \pm 2\pi n$ for any integer $n$, then the signals are not clearly not distinguishable, by the simple trigonometric identity $\sin \theta = \sin \theta + 2\pi$.  Do the math, and you find that the frequencies cannot be separated by more than the sampling rate to be distinguishable.

It's worse, though, because you also have the trigonometric identity $\sin \theta = -\sin \theta + \pi$.  When you consider that the only thing that allows you to distinguish one signal from another is the frequency, that means that if $F_2$ is a mere $\frac{1}{2 T_s}$ away from $F_1$, then $x_1(t)$ cannot be distinguished from $x_2(t)$.

This phenomenon of two distinct signals appearing as one after sampling is called aliasing.  It's pretty much the basis of the Shannon-Nyquist sampling theorem, and the Nyquist rate.  Because if you're sampling at $\frac{1}{T_s}$ and trying to distinguish signals further apart than $\frac{1}{2T_s}$, you just can't.

Hence, the Nyquist rate.