Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer
    
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 4:58
    
@Ross Ah, yeah, typing without thinking. Fixed the couple errors, sorry! –  Emrakul May 24 '13 at 4:59

Your Answer

 
discard

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.