Taylor series in two variables

Solution 1:

Usually $df$ denotes the total derivative. In that case, yes, you are right and $$df=\frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial t}dt.$$

However, in the article, the author is expanding $f$ into its Taylor series. The Taylor series of $f$ (expanded about $(x,t)=(a,b)$ is: $$f(x,t)=f(a,b)+f_x(a,b)\cdot (x-a)+f_t(a,b)\cdot (t-b)+\frac{1}{2}f_{xx}(a,b)\cdot (x-a)^2+$$ $$\frac{1}{2}f_{xt}(a,b)\cdot (x-a)(t-b)+\frac{1}{2}f_{tx}(a,b)\cdot (x-a)(t-b)+ \frac{1}{2}f_{tt}(a,b)\cdot (t-b)^2+\cdots$$

Now, think "dX" means "change in X". So $df=f(x,y)-f(a,b)$, $dx=x-a$, and $dt=t-b$. Thus $$df = f_x dx + f_t dt + \frac{1}{2}f_{xx} dx^2 + \frac{1}{2}f_{xt} dx dt + \frac{1}{2}f_{tx} dx dt + \frac{1}{2}f_{tt} dt^2 + \cdots$$

The total derivative is just the linear approximation of $f$ whereas the Taylor series takes into account higher order terms as well.