Confirmation of differentiability of fixed point with respect to a parameter

Question

$a_{1}=\int_{(r_{1},r_{2})\in A_{1}(a_{2},\beta)} f(r_{1},r_{2})dr_{1}dr_{2}$, $a_{2}=\int_{(r_{1},r_{2})\in A_{2}(a_{1},\beta)} f(r_{1},r_{2})dr_{1}dr_{2}$, where $a_{1},a_{2}\in [0,1]$, $A_{1}(a_{2},\beta)$ is a convex polygon whose boundaries are defined by linear functions of $a_{2}$ and $\beta$, $A_{2}(a_{1},\beta)$ is a convex polygon whose boundaries are defined by linear functions of $a_{1}$ and $\beta$, $f(r_{1},r_{2})$ is probability density function that is continuously differentiable, $\beta$ is a parameter. Example: $A_{1}(a_{2},\beta)=\{(r_{1},r_{2}):r_{2}\leq r_{1}, -\beta \leq r_{1}\leq 2a_{2}\}$, $A_{2}(a_{1},\beta)=\{(r_{1},r_{2}):r_{1}\geq r_{2}+a_{1}\}$, $f(r_{1},r_{2})$ is a uniform density or normal density on $R^2$. I would like to confirm that the implicit function defined by these two equations $a^{*}_{1}(\beta)$ and $a^{*}_{2}(\beta)$ are differentiable in $\beta$. I think they are differentiable, as the right-hand-side of these two equations defined a continuously differentiable mapping $F(a_{1},a_{2},\beta)$, and its Jacobian with respect to $(a_{1},a_{2})$ is invertible, so we can apply the implicit function theorem and conclude that $a^{*}_{1}(\beta)$ and $a^{*}_{2}(\beta)$ are differentiable in $\beta$, right? Thanks.