Taylor series of a function $f(x,t)$ at $(a,b)$ is $$ f(x,t)=f(a,b) +(x-a)f_x(a,b)+(t-b)f_t(a,b) + \cdots .$$ But why $$df=\frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial t}dt + \frac{\partial^2 f}{2\partial x^2}dx^2 + \cdots?$$ This formula is in the 6-th line below Informal derivation. I think that $$df=\frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial t}dt.$$

Thank you very much.

  • 2
    $\begingroup$ I think that's simply wrong. The author should have used "$\Delta$" instead of "$d$" there. Now if only there were a way to suggest improvements to a Wikipedia article ... $\endgroup$ Oct 3, 2011 at 20:19
  • $\begingroup$ @Henning, it may be true. I saw this formula in another place. For example, the second equation $\endgroup$
    – LJR
    Oct 3, 2011 at 20:23

1 Answer 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .