Application of Dominated Convergence Theorem, differentiation and integration commute

Let $f(x,t)$ be a integrable function of $x$. $\frac{\partial f}{\partial t}$ exists and $\frac{\partial f}{\partial t} \leq g$, where $g$ is integrable. Then we have to show $\frac{d}{dt}\int f(x,t)dx=\int \frac{\partial f}{\partial t}dx$.

This is a problem from the book "Measure Theory and Integration" by Gar De Barra. I have taken $x_n \rightarrow x$. Let $F(x)=\int f(x,t) dt$ Then tried to calculate $\frac{F(x_n)-F(x)}{x_n-x}$.Using Dominated convergence theorem I have shown $F(x_n) \rightarrow F(x)$. But while calculating the limit $x_n \rightarrow x$ I was trying to use L'Hospitals Law. But I found that $\lim_{x \rightarrow x_n}\frac{\frac{d}{dx_n} \int (f(x_n,t)-f(x,t))dx}{\frac{d}{dx_n}(x-x_n)}$ ocuurs, and I cannot take the derivative inside the integral.

Thanks in advance for your help.

Let's consider $$ \lim_{h \to 0 } \int \frac {f(x, t + h) - f(x,t)}{h} dx.$$

Since $f$ is differentiable w.r.t. $t$ (for every $x$), the mean value theorem for differentiation gives $$ \frac {f(x, t + h) - f(x,t)}{h} = \frac{\partial f(x, t + \theta(x) h)}{\partial t} $$ where for each $x$, the number $\theta(x)$ is in $[0,1]$.

So $$ \frac {f(x, t + h) - f(x,t)}{h} \leq g(x),$$ which proves that the integral is dominated by an integrable function.

Hence, by the dominated convergence theorem, we have $$ \lim_{h \to 0 } \int \frac {f(x, t + h) - f(x,t)}{h} dx = \int \lim_{h \to 0 } \frac {f(x, t + h) - f(x,t)}{h} dx = \int \frac{\partial f(x,t)}{\partial t}.$$

[Note that the dominated convergence theorem is usually stated for sequences of functions, but this isn't a problem, because we just apply it for every sequence $h_n$ that converges to $0$.]