Assume $f(x), x \in \mathbb{R}$ is a function with a compact support such that its Fourier transform $\hat{f}(\xi)$ has a decay rate $$\hat{f}(\xi) \lesssim \frac{1}{\xi^\gamma + 1}$$ for some $\gamma \ge 1$. Now set $$h(x) = xf(x).$$ Since $f$ has a compact support, $h$ should have similar or better regularity than $f$. Can we now get the following decay estimate of the Fourier transform of $h$ ? $$\hat{h}(\xi) \lesssim \frac{1}{\xi^\gamma + 1}$$ I know now we have $\hat{h}(\xi) = i\partial_\xi \hat{f}(\xi)$, but it seems hard to only use this relation to get the decay estimate.
2 Answers
The answer is positive. Since $f$ has compact support, $g:=\hat{f}$ extends to an entire function of exponential type $\sigma$ with some $\sigma>0$. Then your estimate on the real line and the Phragmen  Lindelof theorem imply that $$\log g(z)\leq \sigma y\gamma\logz,\quad z=x+iy,$$ which gives that $g(z)=O(z^{\gamma}), z\to\infty$ not only on the real line, but also in a horizontal strip $y<2$. Then Cauchy inequality implies that $g'$ also satisfies the same estimate on the real line.
Reference for PhragmenLindelof theorem: B. Ya. Levin, Lectures on entire functions, lecture 6.
Added. Instead of the above inequality, one can write a more precise one, which also follows from PhragmenLindelof: $$\logg(z)\leq \sigmay+\frac{y}{\pi}\int_{\infty}^\infty\frac{\logg(t)}{(xt)^2+y^2}dt.$$ So to obtain an estimate for complex $z$ from the estimate on the real line, one only need to estimate the Poisson integral in the right. This permits to deal with estimates other than $x^{\gamma}$.

$\begingroup$ Many thanks for your answer! It is very helpful! May I ask you one more question? Does the PhragmenLindelof theorem hold for more general functions? For example, if $f(z)$ is an entire function of exponential type and $f(x) \le log(2+x) / (1+x^\gamma)$ on the real axis, can we expect $f(z) \le Ce^{\sigma y}\log(2+z)/(1+z^\gamma)$? Now the estimate has a log in the numerator, will this affect the answer? $\endgroup$– Jacob LuMay 2, 2021 at 3:45
I came up with an answer that is more real analysis. This method works for any $h(x) = m(x)f(x)$ as long as $m(x)$ is smooth and $f(x)$ is compactly supported.
Without loss of generalization, assume the support of $f(x)$ is $[0, 1]$. Set $$g(x) = m(x)\chi(x).$$ Here $\chi(x)$ is a smooth cutoff function such that $\chi(x) = 1$ on $[0,1]$. Hence $g(x)$ is Schwartz and $h(x) = g(x)f(x)$. We then have \begin{equation} \hat{h}(\xi) = \int_{\infty}^\infty \hat{f}(\xi\eta)\hat{g}(\eta)d\eta. \end{equation} For $\xi \gg 1$, we write \begin{equation} \begin{split} \hat{h}(\xi) &= ( \int_{\eta<\xi/2} + \int_{\xi/2 < \eta < 2\xi} + \int_{\eta>2\xi})\hat{f}(\xi\eta)\hat{g}(\eta)d\eta \\ &= T_1 + T_2 + T_3. \end{split} \end{equation} For $T_1$, now $\xi  \eta \sim \xi$, we have $$T_1 \lesssim \frac{1}{\xi^\gamma+1}\int_{\eta<\xi/2}\hat{g}(\eta)d\eta\lesssim \frac{1}{\xi^\gamma+1}.$$ For $T_2$, we have $$T_2 \lesssim \hat{g}(\xi)\int_{\xi/2 < \eta < 2\xi}\frac{1}{1+\xi\eta^\gamma}d\eta \lesssim (1+\xi^{1\gamma})\hat{g}(\xi)$$ which decays much faster than $\hat{f}(\xi)$ since $g$ is Schwartz. For $T_3$, $\xi\eta\sim \eta$, $$T_3 \lesssim \int_{\eta>2\xi}\hat{f}(\eta)\hat{g}(\eta)d\eta \lesssim \frac{1}{\xi^\gamma+1}.$$ In conclusion, it holds that $\hat{h}(\xi) \lesssim \frac{1}{\xi^\gamma+1}.$