Laplace Transform and Fourier Transform of a function I have this transfer function: 
$$
    h(t)= -\frac{1}{16}te^{-2t}
$$
and the Laplace Transform is: 
$$
    H(s) = \frac{-\frac{1}{16}}{(s+2)^{2}}
$$
I know that to find the Fourier Transform, I would just replace s with jw
$$
H(j\omega) = \frac{-\frac{1}{16}}{(j\omega+2)^{2}}
$$
But, the Fourier Transform table says: 
$$
tf(t) \Leftrightarrow j\frac{d}{d\omega}\hat{f}(\omega)
$$
And the Fourier transform of $$ e^{-2t} \Leftrightarrow \frac{1}{j\omega + 2} $$
$$ -\frac{1}{16}te^{-2t} \Leftrightarrow  \frac{j\frac{1}{16}}{(j\omega + 2)^{2}} $$
So, the final answer is not the same as $H(j\omega)$!
Do you see any thing wrong here? Please guide me!

When you differentiate, there is another $j$ in the numerator giving $j^2$.
Also, your $h(t)$ should be
$$
h(t)= -\frac{1}{16}te^{-2t} ~{\mathbf u(t)}
$$
