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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?

student lee
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2 Answers2

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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.

Ross Millikan
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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.

  • We are taking the derivative with respect to $x$, not $n$. There is no derivative of $n+1$ Also no function is differentiable at a discontinuity. You are correct that a large class of functions, continuous in both $x$ and $n$ satisfy this. – Ross Millikan May 24 '13 at 04:58
  • @Ross Ah, yeah, typing without thinking. Fixed the couple errors, sorry! –  May 24 '13 at 04:59