$\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$ $\,$Let $U \subseteq C$ be open and let $\gamma: [a,b] \rightarrow U$ be a $C^1$ curve if $f: U \rightarrow R \, \, and \, \, f \, \in\, C^1(U)$ Then we have the following. $$f(\gamma(b))-f(\gamma(a)) = \int_{a}^{b}(\frac{\partial f}{\partial x}(\gamma(t)) \cdot \frac{d{\gamma_1}}{dt} + \frac{\partial f}{\partial y}(\gamma(t)) \cdot \frac{d\gamma_2}{dt})$$
Lemma: Looking at the RHS side of our proposition I noticed that "Feynman's Integration Trick" stated in 2.) can be applied as follows in 3.)
2.) Feynman's Integration Trick is stated as follows:
$$\frac{d}{dx}\int_{Y}f(x,y)dy=\int_{Y}\frac{\partial}{\partial x}f(x,y)dy$$
Applying our Lemma to our original proposition I observed the following occurs when applied.
$$\frac{d}{dx}\int_{a}^{b}(\frac{\partial f}{\partial x}(\gamma(t)) \cdot \frac{d{\gamma_1}}{dt} + \frac{\partial f}{\partial y}(\gamma(t)) \cdot \frac{d\gamma_2}{dt})=\int_{a}^{b}\frac{\partial}{\partial x}(\frac{\partial f}{\partial x}(\gamma(t)) \cdot \frac{d{\gamma_1}}{dt} + \frac{\partial f}{\partial x}(\gamma(t)) \cdot \frac{d\gamma_2}{dt})$$
Remark: The reason that Feynman's Integration Trick can be applied is due to the fact since our functions are $C^1$ because those functions lie on a closed interval.
From my conclusion in 3.) I'm having trouble applying the lemma in this particular situation. My initial question is how to apply the technique within the lemma further in 3.) to reach a conclusion.