property of function Let $f _n (x)=x ^n$  . 
If I want to get $f_{n+1}'(x)$ , 
firstly I find $f _{n+1} (x)=x^ {n+1}$   and next differentiate $f _{n+1} (x)=x^ {n+1}$  , 
I obtain $f _ {n+1}' (x)=(n+1)x^ n $ . 
But in other ways, firstly differentiate $f _n (x)$ , obtain $f  _n '(x)=nx^ {n−1} $ , and substitute $n+1$  for $n$  in $f  _n '(x)=nx^ {n−1}$  , I get $f_  {n+1}' (x)=(n+1)x ^n$   too. 
I think second ways is not definition of $f ′ _{n+1} (x)$ . 
Is there any function which satisfying first ways is not equal to second ways? 
 A: Many functions satisfy this, but only because you are viewing $n$ as a fixed parameter.  You really have a function $f(x,n)$ and are asking does $\frac {\partial f(x,n)}{\partial x}|_{n \to n+1}=\frac {\partial f(x,n+1)}{\partial x}$ or do partials and substitution commute?  So let me define $f_n(x)= \begin {cases} x&n=1\\5x&n=2 \end {cases}$ and it fails.  Good for you to be thinking about this.
A: As long as you stick to real, continuous functions, no. Here's the reasoning:


*

*In the first example you provided, you find the next term, then take its derivative. You end up with the derivative of the $(n+1)$th term.

*In the second example you provided, you find the general derivative term, then plug in $n+1$. You end up with the derivative of the $(n+1)$th term. 


To think about it another way:
$$f_n{x}\to f_{n+1}{x}\to f'_{n+1}{x}$$
$$f_n{x}\to f'_n{x}\to f'_{n+1}{x}$$
Both get you to the same place. There's no reason why this would change. However, for discontinuous functions, it could very well be different.
