Taylor series expansion of function whose argument has additional dependency

The Taylor series expansion of a function, $$f(x)$$, around point "a" is

$$f(x) = f(a) + \frac{x-a}{1!}\frac{\partial f(a)}{\partial x} + \frac{(x-a)^2}{2!}\frac{\partial^2 f(a)}{\partial x^2} + \ldots$$

which is well known (wikipedia link).

My question is, what happens if "x" is a function of "t"? Is there a Taylor series expression, similar to that above? Would it be valid for all time?

Assuming there is a valid expansion...

What if the function "f" is a function of "x" and "t" (in this case "x" does not depend on "t")? I believe this is just a multi-variable Taylor expansion as found on the wikipedia link above. How would the multi-variable series expansion differ from that above in my question...where "x" is a function of "t"?

Developed around $$t=a$$, you should have $$f(x(t))=f(x(a))+x'(a) f'(x(a))(t-a) +\frac{1}{2!} \left(x'(a)^2 f''(x(a))+x''(a) f'(x(a))\right)(t-a)^2+\frac{1}{3!} \left(f'''(x(a)) x'(a)^3+3 x'(a) x''(a) f''(x(a))+x'''(a) f'(x(a))\right)(t-a)^3+O\left((t-a)^4\right)$$ For the next case of $$f(x(t),t)$$, it starts to be really tedious, involving the partial derivatives of $$f$$ with respect of $$x$$ and $$t$$ but it is doable.
• The chain rule stuff, with the derivatives, makes sense. That's kind of what I thought, but I wasn't sure if the $(x-a)$ simply translated to $(t-a)$ due to x's dependency on t. Can you maybe comment/show/point to reasoning as to how that part manifests? Also, would this series expansion hold for all time, despite the variability of x? – ThatsRightJack Jan 6 at 11:38
• As for $f(x(t),t)$...I'm going to edit that part out of the question because it is asking too much I think. – ThatsRightJack Jan 6 at 11:41