Fourier Transform with both Time Delay and Frequency Shift I know that the Fourier transform of a function with time delay can be written as: $$\mathscr{F}\big\{x(t-t_0)\big\}=X(f)e^{-j2\pi f t_0}$$

The Fourier transform of a function with frequency shift can also be written as: $$\mathscr{F}\Big\{x(t)e^{j2\pi f_0 t}\Big\}=X(f-f_0)$$

So what if we have both shift and delay at the time domain, what will be the result in the frequency domain? E.g.:

$$\mathscr{F}\Big\{x(t-t_0)e^{j2\pi f_0 (t-t_0)}\Big\}$$

Will the result be: $$X(f-f_0)e^{-j 2 \pi  f (t-t_0)}$$

Also what will be the result of: 

$$\mathscr{F}\Big\{x(t-t_0)e^{j2\pi f_0 t)}\Big\}$$

Is there an order to apply these properties?
If you are ever unsure, just go back to the definition and work out the Fourier Transform property for the specific situation:

$$\begin{align*}\mathscr{F}\left\{x\left(t-t_0\right)e^{j2\pi f_0\left(t-t_0\right)}\right\} &amp;= \int_{-\infty}^\infty x\left(t-t_0\right)e^{j2\pi f_0\left(t-t_0\right)} e^{-j2\pi f t}dt\\
\\
&amp;= \int_{-\infty}^\infty x\left(\tau\right)e^{j2\pi f_0\tau} e^{-j2\pi f \left(\tau+t_0\right)}d\tau \\
\\
&amp;= e^{-j2\pi ft_0}\int_{-\infty}^\infty x\left(\tau\right)e^{j2\pi f_0\tau} e^{-j2\pi f \tau}d\tau \\
\\
&amp;= e^{-j2\pi ft_0}\int_{-\infty}^\infty x\left(\tau\right) e^{-j2\pi (f-f_0) \tau}d\tau \\
\\
&amp;= e^{-j2\pi ft_0} X(f-f_0)\\
\\
\end{align*}$$

$$\begin{align*}\mathscr{F}\left\{x\left(t-t_0\right)e^{j2\pi f_0 t}\right\} &amp;= \int_{-\infty}^\infty x\left(t-t_0\right)e^{j2\pi f_0t} e^{-j2\pi f t}dt\\
\\
&amp;= \int_{-\infty}^\infty x\left(\tau\right)e^{j2\pi f_0(\tau+t_0)} e^{-j2\pi f \left(\tau+t_0\right)}d\tau \\
\\
&amp;= e^{-j2\pi (f-f_0)t_0}\int_{-\infty}^\infty x\left(\tau\right)e^{j2\pi f_0\tau} e^{-j2\pi f \tau}d\tau \\
\\
&amp;= e^{-j2\pi (f-f_0)t_0}\int_{-\infty}^\infty x\left(\tau\right) e^{-j2\pi (f-f_0) \tau}d\tau \\
\\
&amp;= e^{-j2\pi (f-f_0)t_0} X(f-f_0)\\
\\
\end{align*}$$