Fourier transform of $\phi(x) = \sinh (x) \delta[(x-\log2)\sinh(x)]$ and $\phi(x)=\dfrac{d}{dx}[xH(x-1)]$

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I have to find the Fourier transform of those functions: $$\phi(x) = \sinh (x) \delta[(x-\log2)\sinh(x)]$$ $$\phi(x)=\dfrac{d}{dx}[xH(x-1)]$$ where $\delta$ is Dirac delta and $H$ is Heaviside function ($H = 1$ if $x\ge0$ and $H=0$ if $x<0$).

I tried using some properties of Dirac delta and Fourier transform, such as $\delta[g]=\sum_{i}\dfrac{\delta(x-a_i)}{|g'(a_i)|}$, but I can't get anything good.

Can you help me? Thanks in advance.

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You can use what you have: $$\delta[g]=\sum_{i}\dfrac{\delta(x-a_i)}{|g'(a_i)|}.$$ Say that $$g(x)= (x-\ln 2)\sinh(x)$$ so its two roots are $0$ and $\ln 2$.

The derivative of $g$ is $$g'(x)=\sinh (x)+(x-\ln2)\cosh(x),$$ so $|g'(0)|=\ln 2$ and $|g'(\ln 2)|=3/4.$

Finally we can use the property above to obtain

$$\delta[g](x)=\frac{\delta(x)}{\ln2}+\frac{4\delta(x-\ln 2)}{3}. $$

Now the Fourier transform should be no big deal.