Lawrence Dresner says this: (p. 10, Applications of Lie's Theory of Ordinary and Partial Differential Equations)
Assume you have two infinitesimal group transformations: $$x'=x+\varepsilon(\lambda - \lambda_o)$$ $$y'=y+\eta(\lambda - \lambda_o)$$
Then $$dx'=dx+d\varepsilon(\lambda - \lambda_o)$$ $$dy'=dy+d\eta(\lambda - \lambda_o)$$ Upon dividing these equations, we find
$$\dot{y}'\equiv \frac{dy'}{dx'}=\bigg[\dot{y}+\frac{d\eta}{dx}(\lambda - \lambda_o)\bigg]\bigg[1+\frac{d\varepsilon}{dx}(\lambda - \lambda_o)\bigg]^{-1}=\dot{y}+\bigg(\frac{d\eta}{dx}-\dot{y} \frac{d\varepsilon}{dx}\bigg)(\lambda-\lambda_o)$$ The important result here is that the expression $\frac{d\eta}{dx}-\dot{y} \frac{d\varepsilon}{dx}$ is the infinitesimal transformation for the once-extended group. In fact, there's a recursive formula for these.
I understand that $$\frac{dy'}{dx'}=\frac{\frac{dy'}{dx}}{\frac{dx'}{dx}}$$
but what I don't understand is the last step. I know that this result works. For example, suppose you have a simple Lie group:
$$G(x,y)=(\lambda x,\lambda^\beta y)\lambda_o=1$$ such that $x'=\lambda x$ and $y'=\lambda^\beta y$. From this you can also see that
$$\dot{y}'=\frac{dy'}{dx'}=\frac{\lambda^\beta y}{\lambda x}=\lambda^{\beta -1}\dot{y}$$
Finding the infinitesimal transformations for this group:
$$\frac{\partial x'}{\partial \lambda}\bigg|_{\lambda_o =1}=x=\varepsilon$$
$$\frac{\partial y'}{\partial \lambda}\bigg|_{\lambda_o =1}=\beta y=\eta$$
$$\frac{\partial \dot{y}'}{\partial \lambda}\bigg|_{\lambda_o =1}=(\beta-1)\dot{y}=\eta_1$$
However, if you take Dresner's statement at face value, you can calculate $\eta_1$ directly from $\varepsilon$ and $\eta$.
$$\frac{d\eta}{dx}-\dot{y}\frac{d\varepsilon}{dx}=\beta \dot{y} -\dot{y}(1) = (\beta -1)\dot{y}$$
So.....it works! I just don't understand HOW. Please help!