Differentiating under the integral sign on a manifold

Question

Sorry about the previous version of the question. I will try to be more precise and more to the point Given a smooth, compact, oriented manifold $M^1 $ with $\partial M=\emptyset$ , a 1-form $\lambda$ and the flow of a smooth vector field $\theta_t$ I have the function $$t\mapsto \intop_M\theta_t^*\lambda$$ I want to differentiate with respect to $t$ under the integral sign. Now, by definition of integration on a manifold we have (with $\{f_\alpha\}_\alpha$ a partition of unity) $$\intop_M\theta_t^*\lambda = \intop_M\lambda\circ\theta_t=\sum_{\alpha\in I} \intop _{\varphi_\alpha(U_\alpha)} f_\alpha\lambda_1(\theta_t(x))d(\theta_t(x))$$ And here is the pickle. I have $t$ appearing in the integration variable. So I would like to say that since the integrand is smooth I can just do (for fixed $\alpha$) $$\frac d {dt} \intop _{\varphi_\alpha(U_\alpha)} f_\alpha\lambda_1(\theta_t(x))d(\theta_t(x)) = \intop _{\varphi_\alpha(U_\alpha)} f_\alpha \frac \partial {\partial t} \lambda_1(\theta_t(x))d(\theta_t(x))$$ and then regroup everything together to obtain $$\intop_M \frac \partial {\partial t} \theta_t^*\lambda$$ which happens to be integration of the Lie derivative of $\theta_t^*\lambda$ The question is whether this is true and if so, why?