Let $\phi:[a,b]\times[c,b]\to \mathbb{R}$ be continuous. Define $h:[c,d]\to \mathbb{R},h(t)=\int_{a}^{b}\phi(s,t)ds$. Assume that $\frac{\partial\phi}{\partial t}$ exists and is continuous on $[a, b]\times[c, b]$. Prove that $h'(t)=\int_{a}^{b}\frac{\partial\phi}{\partial t}ds$.
I have tried to prove it by first differentiating $h(t)$ w.r.t $t$ but it shouldn't be that easy. Is it okay to just differentiate on $h(t)$ w.r.t $t$ directly? Or how to prove it?
Although one can use measure theory (e.g. the dominated convergence theorem) to prove this, there is however a direct approach since we are dealing with a continuous function. Let $t_0 \in [c,d]$ and $\varepsilon>0$. Since $\displaystyle\frac{\partial\phi}{\partial t}(s,t_0)$ exists, then for each $s \in [a,b]$ there is a $\delta(t_0,s,\varepsilon)>0$ such that $$ \left|\frac{\partial\phi}{\partial t}(s,t_0)-\frac{\phi(s,t)-\phi(s,t_0)}{t-t_0}\right|< \frac{\varepsilon}{b-a} \quad \forall (s,t) \in [a,b]\times[c,d],\ |t_0-t|<\delta(t_0,s,\varepsilon). $$ Since $$ [a,b]\times[c,d] \subset \bigcup_{(s,t) \in [a,b]\times[c,d]}(s-\varepsilon,s+\varepsilon)\times(t-\delta(t,s,\varepsilon),t+\delta(t,s,\varepsilon)) $$ and $[a,b]\times[c,d]$ is compact, therefore there exist $$ (s_1,t_1),\ldots, (s_n,t_n) \in [a,b]\times[c,d] $$ such that $$ [a,b]\times[c,d] \subset \bigcup_{i=1}^n(s_i-\varepsilon,s_i+\varepsilon)\times(t_i-\delta(t_i,s_i,\varepsilon),t_i+\delta(t_i,s_i,\varepsilon)). $$ Choosing $$ \delta_0(\varepsilon)=\min_{1 \le i \le n}\delta(t_i,s_i,\varepsilon), $$ we have for every $t \in [c,d]$ with $|t_0-t|< \delta_0(\varepsilon)$ \begin{eqnarray} \left|\int_a^b\frac{\partial\phi}{\partial t}(s,t_0)ds-\frac{h(t)-h(t_0)}{t-t_0}\right|&=&\left|\int_a^b\left[\frac{\partial\phi}{\partial t}(s,t_0)-\frac{\phi(s,t)-\phi(s,t_0)}{t-t_0}\right]ds\right|\\ &\le&\int_a^b\left|\frac{\partial\phi}{\partial t}(s,t_0)-\frac{\phi(s,t)-\phi(s,t_0)}{t-t_0}\right|ds\\ &\le&\int_a^b\frac{\varepsilon}{b-a}ds=\varepsilon, \end{eqnarray} i.e. $h$ is differentiable at $t_0 \in [c,d]$ and $$ h'(t_0)=\int_a^b\frac{\partial\phi}{\partial t}(s,t_0)ds. $$
Since $\phi$ is continuous and smooth (its derivative is also continuous), and since integration domain is fixed, then you can exchange integral with derivative operator.
\begin{equation} \nonumber \begin{array}{lcl} h'(t) & = & \frac{\partial h}{\partial t} = \\ & = & \frac{\partial}{\partial t} \int_{a}^{b}\phi(s,t)ds = \\ & = & \int_{a}^{b} \frac{\partial \phi(s,t)}{\partial t}ds & \end{array} \end{equation}
This is the Leibniz integral rule.
Actually,it's a corollary of the Lebesgue dominated convergence theorem.The exchange is availbale here because $\frac{\partial \phi(s,t)}{\partial t}$ is dominated by a integrable function i.e. it's upper-bound in $[a,b]\times[c,b]$.
