I have the next: Let be $U \subset \mathbb{R}^{m}$ open, $\phi: U\times [a,b] \rightarrow{\mathbb{R}^{n}}$ with partial derivate continuos $\partial_{1}: U\times [a,b] \rightarrow{T(\mathbb{R}^{m}, \mathbb{R}^{n})} $ and $\alpha : U \rightarrow{[a,b]}$ of class $C^{1}$. If $f: U \rightarrow{\mathbb{R}^{n}}$ definited by : $f(x) = \int _{a}^{\alpha(x)} \phi(x,t)dt$, $x \in U$, then $f \in C^{1}$. Find $f^{\prime}(x)h$.
My attempt: Let be $\xi: U\times [a,b] \rightarrow{\mathbb{R}^{n}}$ definited by $\xi (x,u)= \int _{a}^{u} \phi(x,t)dt$, then we have the next: $\frac{\partial \xi(x,u)}{\partial x_{i}} = \int _{a}^{u} \partial _1\phi(x,u)dt$, which is continuous, because $\partial_{1}$ is continuos in the points x for all $u \in [a,b]$, and how $\frac{\partial \xi(x,u)}{\partial u} = \phi(x,u)$ is continuos, so $ \xi: U\times [a,b] \rightarrow{\mathbb{R}^{n}}$ is the class $C_{1}$.
Then as $f(x) = \xi(x, \alpha(x))$ we have $f(x)$ is differentiable $C^{1}$. Now, how $\xi^{\prime}(x,u) = \int_{a}^{u}\partial_{1}\phi(x,u)dt$ we have to: $f^{\prime}(x) = \xi^{\prime}(x,\alpha(x))\begin{pmatrix}{I}\\{\nabla(\alpha(x))}\end{pmatrix}$.
the questions are:
- ¿Is it correct?
- If i wanted to see explicitly the the entries of $f^{\prime}(x)$. (The truth is I don't know how to do it, since I don't know what the matrix of $\xi^{\prime}(x, \alpha(x))$ in terms of its partial derivatives, I complicate the sign of the integra.) ¿How can I do it? in case it can be done.