Fourier Transform Simple Problem

I have a simple question.

x(t) = derivative of [sin(t)/(pi*t) convolution sin(2t)/(pi*t)] dt.

I need the Fourier Transform of x(t).

Let's call the second part h(t). So, h(t) = sin(2t)/pi*t. Using tables, the Fourier Transform of h(t) would be
1, if |ω| < 2
0, if |ω| > 2
right?

Calling the other part g(t) = sin(t)/pi*t, the Fourier Transform would be
1, if |ω| < 1
0, if |ω| > 1
right?

So, I know I have to use the convolution property, which states that:
g(t) convolution h(t) = G(jω) times H(jω).... just the multiplication of the Fourier Transforms.

Can somebody help me with this part? This multiplication would result in what?

If I get that, then I just need to apply the derivative formula for the Fourier Transform to get to my answer. In other words, X(jω) = jω * G(jω) * H(jω), right?

Help please!