I wrote an article on analysis recently and I included the following relevant result (with proof) in the article; I hope it is helpful:
Theorem Let $f\in L^1(\mathbb{R}^n)\cap L^p(\mathbb{R}^n)$ for some $1\leq p \leq \infty$. Also, let $g\in L^1(\mathbb{R}^n)$ be a function all of whose partial derivatives of the first order exist and are such that $\frac{\partial g}{\partial x_i}$ is bounded on $\mathbb{R}^n$ for all $1\leq i\leq n$. We conclude that the partial derivatives of the convolution $f\ast g$ of the first order exist on $\mathbb{R}^n$. In fact, $\frac{\partial (f\ast g)}{\partial x_i}=f\ast (\frac{\partial g}{\partial x_i})$ for all $1\leq i\leq n$.
Proof. First note that the convolution $f\ast g\in L^1(\mathbb{R}^n)\cap L^p(\mathbb{R}^n)$ by Minkowski's inquality
and is therefore finite (and well-defined) a.e. Let us fix $1\leq i\leq n$.
Note that
$\frac{\left(f\ast g\right)\left(x+he_i\right)-\left(f\ast g\right)(x)}{h} - \left(f\ast \left(\frac{\partial g}{\partial x_i}\right)\right)\left(x\right)$
$= \int_{\mathbb{R}^{n}} f\left(y\right)\left[\frac{g\left(x-y+he_i\right)-g\left(x-y\right)}{h} - \left(\frac{\partial g}{\partial x_i}\right)\left(x-y\right) \right]dy$
In particular,
$\frac{\partial \left(f\ast g\right)}{\partial x_i}\left(x\right)$
$=\lim_{h\to 0} \int_{\mathbb{R}^{n}} f\left(y\right)\left[\frac{g\left(x-y+he_i\right)-g\left(x-y\right)}{h}\right]dy$
$= \int_{\mathbb{R}^{n}} \left[\lim_{h\to 0} f\left(y\right)\left[\frac{g\left(x-y+he_i\right)-g\left(x-y\right)}{h}\right]\right]dy$
$= \int_{\mathbb{R}^n} f\left(y\right)\frac{\partial g}{\partial x_i}\left(x-y\right) dy$
$= \left(f\ast \frac{\partial g}{\partial x_i}\right)\left(x\right)$
We will justify this computation using the
Lebesgue dominated convergence theorem. In particular, we will show that if $x\in \mathbb{R}^n$ is fixed,
the expression $\left|\frac{g\left(x-y+he_i\right)-g\left(x-y\right)}{h} - \left(\frac{\partial g}{\partial x_i}\right)\left(x-y\right)\right|$ is bounded by
an $L^1(f)$ function in $y$ for all $h>0$ sufficiently small. (Let us recall that $L^1(f)$ is the $L^1$ space
associated to the complex measure $\mu_f$ defined by $\mu_f(E)=\int_{E} f$ for every measurable $E\subseteq \mathbb{R}^n$.
Clearly, every constant function is in $L^1(f)$.)
However, this is an easy consequence of the mean value theorem: we know that there exists
$\delta>0$ such that $0
$\left|\left[\frac{g\left(x-y+he_i\right)-g\left(x-y\right)}{h}\right] -
\left(\frac{\partial g}{\partial x_i}\right)\left(x-y\right)\right|$
$\leq 2\sup_{c\in\mathbb{R}^n}\left|\frac{\partial g}{\partial x_i}\left(c\right)\right|$
and the result now follows from the hypotheses. Q.E.D.