The correct answer is $\boxed{C. jfX\left( f \right)}$.
The Fourier transform of a function $x(t)$ is defined as
$$X(f) = \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} dt$$
The derivative of a function can be written in terms of the Fourier transform as
$$\frac{dx(t)}{dt} = -j2\pi f \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} dt = -j2\pi f X(f)$$
Therefore, the Fourier transform of $\frac{dx(t)}{dt}$ is $jfX(f)$.
Option A is incorrect because it is the original function $x(t)$, not its derivative. Option B is incorrect because it is the Fourier transform of the function $2\pi ft$, which is not the derivative of $x(t)$. Option D is incorrect because it is the inverse Fourier transform of $jfX(f)$, which is not the derivative of $x(t)$.