Differentiating under the integral sign in $n$ variables to show $\partial^{\alpha}(f*g)=f*\partial^{\alpha}g$

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As an easy corollary of Dominated Convergence theorem, Folland gives a criterion for differentiating under the integral sign and uses it to prove Proposition $8.10$: Folland Theorem 2.27

Proposition $8.10:$ If $f \in L^1$, $g \in C^k$ and $\partial^{\alpha}g$ is bounded for all multi-indices $\alpha$ with $|\alpha| \leq k$, then the convolution $f*g \in C^k$ and $\partial^{\alpha}(f*g)=f*\partial^{\alpha}g$.

As proof, Folland simply states this is clear from Theorem $2.27$ as stated above.

Attempt: We have $f*g(x)=\int f(y)g(x-y)dy$.

Letting $G:\mathbb{R^n} \times [a,b] \rightarrow \mathbb{C}$, where $G(y,x_1)=f(y)g(x-y)$ where $x_i$ is fixed for all $i>1$.

Then we see hypotheses in Theorem $2.27.$b are satisfied and we get

$\frac{\partial}{\partial x_1} f*g(x)=\int f(y)\frac{\partial}{\partial x_1}g(x-y) dy=f*\frac{\partial}{\partial x_1}g $.

Now we simply induct on $|\alpha|$ to get the desired result.

Is this how we justify differentiating under the integral sign in $n$ variables?