Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 ... – Henning Makholm Oct 3 '11 at 20:19
@Henning, it may be true. I saw this formula in another place. For example, the second equation – LJR Oct 3 '11 at 20:23
up vote 4 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.