I have edited my post once more. Assume $\eta(x,t)$ is evaluate first at x=0 and then at $t=t-\frac{x}{1+v}$ where $v$ is a constant. This gives us $$\eta(0,t-\frac{x}{1+v})$$ now i wish to take $\frac{d}{dx}$ of the above so: $$\frac{d}{dx}\eta(\underbrace{0}_\textrm{a},\underbrace{t-\frac{x}{1+v}}_\textrm{b})=\frac{d\eta(a,b)}{da}\frac{da}{dx}+\frac{d\eta(a,b)}{db}\frac{db}{dx}$$ since $\frac{da}{dx}=0$ and $\frac{db}{dx}=-\frac{1}{1+v}$ we have $$=\frac{d\eta(a,b)}{db}(-\frac{1}{1+v})$$ but what is $\frac{d}{db}?$,
Is there anything that prevents me from wtitting above as: $$=\frac{d\eta(a,b)}{dt}\frac{dt}{db}(-\frac{1}{1+v})=\eta_t(a,b)(1)(-\frac{1}{1+v})$$
or for that matter: $$=\frac{d\eta(a,b)}{dx}\frac{dx}{db}(-\frac{1}{1+v})=\eta_x(a,b)(-(1+v))(-\frac{1}{1+v})=\eta_x(a,b)$$ If this is incorrect, where am I confused? Any help is greatly appreciated. Thank You,