Suppose I have a two functions $g,f:\mathbb{R_+^2}\rightarrow\mathbb{R_+}$. Both are functions in two arguments $x_1$ and $x_2$. Define a new function $H:$$$H(x_1,x_2)=f(x_1,x_2)+g(x_1,x_2)$$
I know that $\frac{\partial f(x_1,x_2)}{\partial x_1}=\frac{\partial g(x_1,x_2)}{\partial x_2}=1$. What is the sign of $\frac{\partial H(x_1,x_2)}{\partial x_1}$?
Differentiating $H(x_1,x_2)$ wrt to $x_1$ yields $$1+\frac{\partial g(x_1,x_2)}{\partial x_1}$$
We have that $$\frac{\partial H(x_1,x_2)}{\partial x_1}>0 \iff 1+\frac{\partial g(x_1,x_2)}{\partial x_1}>0$$
Is this correct?
Yes, assuming $g$ is differentiable wrt $x_1$ it's true that $\frac{\partial H(x_1,x_2)}{\partial x_1}= 1+\frac{\partial g(x_1,x_2)}{\partial x_1}$.
One interesting thing to note is (pretending you can extend the domains to $0$) that $f,g$ map to $R_+$, so $$g(x_1,x_2)=g(0,0)+\int_0^{x_1}\frac{\partial g(x,0)}{\partial x}dx+\int_0^{x_2}\frac{\partial g(x_1,x)}{\partial x}dx$$ $$=g(0,0)+\int_0^{x_1}\frac{\partial g(x,0)}{\partial x}dx+x_2$$, so
$$\frac{\partial g(x_1,x_2)}{\partial x_1}=\frac{\partial g(x_1,0)}{\partial x_1}$$ doesn't depend on $x_2$.