This is actually a very old question that now I have to face it again and look for answer of it. Suppose $f:\mathbb{R}^n\to \mathbb{R}, f(x_1,x_2,...,x_n)=y$ is a function. What is the difference between:

  • $\frac{\partial f}{\partial x_i}$
  • $\frac{df}{dx_i}$ if this means anything at all?

I am reading this book and the following passage is part of the book:

...it would be appropriate to introduce a scaled time $\tau$ via

$$\tau=\epsilon^2 t$$

and regard $u$ as depending both on $t$ and $\tau$, and having no explicit dependence on $\epsilon$; $t$ and $\tau$ will be treated as mutually independent. Correspondingly, the time differentiation should be transformed as

$$\frac{d}{dt}\to\frac{\partial}{\partial t}+\epsilon^2 \frac{\partial}{\partial \tau} $$

Where $u=X-X_0$ and $X_0$ is a stable answer for following differential equation: $$\frac{dX}{dt}=F(X)$$

  • $\begingroup$ @CameronWilliams Thanks. oops! I made a mistake in the denominator of the second one! It should be $\frac{df}{dx_1}$. Is that meaningful then? $\endgroup$
    – Cupitor
    Feb 18, 2014 at 19:47
  • $\begingroup$ @CameronWilliams, but if this limit is $\frac{df}{dx_i}$ what exactly is the $\frac{\partial f}{\partial x_i}$? $\endgroup$
    – Cupitor
    Feb 18, 2014 at 19:54
  • $\begingroup$ @CameronWilliams, I am rally sorry I don't have my glasses and that was the reason I couldn't differentiate between $i$ and $1$. So based on what you are saying $\frac{df}{dx_i}\equiv \frac{\partial f}{\partial x_i}$ ?? $\endgroup$
    – Cupitor
    Feb 18, 2014 at 19:56
  • $\begingroup$ Oh sorry. I didn't even realize you were using $d$ and I was too. $\frac{df}{dx_i}$ means nothing. All of my responses should have said $\frac{\partial f}{\partial x_i}$. $\endgroup$ Feb 18, 2014 at 19:56
  • $\begingroup$ But then there is this book on Dynamical Systems that I am reading and actually approximates one of these based on the other. Let me edit my question. Thanks a lot. $\endgroup$
    – Cupitor
    Feb 18, 2014 at 19:57

1 Answer 1


$$\frac{\partial f}{\partial x_i}$$

Is a partial derivate so you suppose that all variables except $x_i$ are constant (you can this $f$ as a single variable function).


Is a total derivate (all variables may vary). So you have to apply the chain rule and you get:

$$\frac{df}{dx_i}=\sum_j \frac{\partial f}{\partial x_j}\frac{dx_j}{dx_i}$$

  • $\begingroup$ Thank you. Vote up. It kinda make sense but still I don't see how the limit base definition of this one is gonna look like? $\endgroup$
    – Cupitor
    Feb 18, 2014 at 20:49

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