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Lets say we have a function $f(x)$ that has a derivative at point $a$. Can we prove that the function $x\cdot f(x)$ has also a derivative at point $a$? If this is not true, can anybody give an example that shows this ...

Git Gud
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stevensitas
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  • @stevensitasIt is true because the product of differentiable functions is a differentiable function and both $x\mapsto x$ and $f$ are differentiable. Innit, bruv? – Git Gud May 11 '13 at 21:14
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    Can you provide some context? What do you know? I'm guessing you're supposed to prove it by definition. Do you know the definition of derivative of a function? – Git Gud May 11 '13 at 21:20
  • I know that the product of differentiable functions is a differentiable function. But in my question the function f(x) is given differentiable at only one point (a). Could this be a problem? – stevensitas May 11 '13 at 22:46
  • You need to understand that saying: "if $g,h$ are differentiable, then so is $gh$" means "if $g,h$ are differentiable on a certain set, then so is $gh$", which in turns means that "if $g,h$ are differentiable on $a$ for every $a$ on a certain set, then $gh$ is differentiable on $a$ for every $a$ on that set". So you see, what you wanted follows imediately from the meaning of those sentences. – Git Gud May 11 '13 at 22:50
  • So the above works not only on sets of type (a,b) but also on the set [a,a] on the real axes? – stevensitas May 11 '13 at 22:57
  • So I have a function f(x) in the following way: (1) f(x)=x for x=a and f(x)="a non differentiable function" for x not equal to a. So f(x) can only be differentiable at point a AND No Where else (near or far a). Am I missing something? – stevensitas May 11 '13 at 23:20
  • I'm off now. That question should be asked seperately with a link to this question. Plus by doing that it will get more views. I told you something wrong on a comment above (which I already deleted). Check this. See you. – Git Gud May 11 '13 at 23:22

3 Answers3

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To avoid using anything but the definition:

Let $g(x)=xf(x)$ $$\lim_{h\to 0} \frac{g(x+h)-g(x)}{h} = \lim_{h\to 0} \frac{(x+h)f(x+h)-xf(x)}{h} = \lim_{h\to 0} \frac{xf(x+h)-xf(x)+hf(x+h)}{h} = x\lim_{h\to 0} \frac{f(x+h)-f(x)}{h} +\lim_{h\to 0}f(x+h) = xf^\prime(x)+f(x)$$

Alfonso Fernandez
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If you don't know the prodouct rule (yet), you can calculate explitly for $g(x)=xf(x)$ $$\begin{align}g'(a)&=\lim_{h\to 0}\frac{g(a+h)-g(a)}{h}\\&=\lim_{h\to 0}\frac{(a+h)f(a+h)-af(a)}{h}\\&=\lim_{h\to 0}\frac{af(a+h)-af(a)+hf(a+h)}{h}\\&=\lim_{h\to 0}\left(a\cdot\frac{f(a+h)-f(a)}{h}+f(a+h)\right)\\ &=a\lim_{h\to0}\frac{f(a+h)-f(a)}{h}+\lim_{h\to0}f(a+h)\\&=af'(a)+f(a).\end{align}$$ Make yourself clear that each step is justified.

Hagen von Eitzen
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Hint: Let $h(x) = x\cdot f(x)$. By the product rule, what is $h'(a)$?

Pedro
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Alex Wertheim
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  • You're assuming differentiability from the get go. – Git Gud May 11 '13 at 21:15
  • Only of $f(x)$, which is assumed to begin with, right? Or am I being careless? – Alex Wertheim May 11 '13 at 21:16
  • No, not only on $f$. You can't talk about $h'$ before estabilishing $h$ is differentiable. – Git Gud May 11 '13 at 21:17
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    @GitGud Well, the product rule says precisely, "let $f,g$ be differentiable. Then $f\cdot g$ is differentiable and $(f\cdot g)'=f'g+fg'$." This looks fine to me. Or do you want the guy to prove $x\mapsto x$ is differentiable? – Pedro May 11 '13 at 21:20
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    Forgive me if I'm being dense (or worse, circular), but as you say, you only need to know that the the two functions in the product are differentiable to conclude that the the product itself is differentiable. If one were to rigorously evaluate the derivative $h'$, one would end up with limits that are contingent on the differentiability of $f(x)$ and $x$. – Alex Wertheim May 11 '13 at 21:20
  • @AWertheim Yes, I did say that. But most likely the OP doesn't know that yet, that's why I left it as a comment. – Git Gud May 11 '13 at 21:21
  • Fair enough, reasonable objections - I didn't mean to be glib in my answer, but I guess I should have explicitly acknowledged the point made in your comment. Thanks for the edit @PeterTamaroff! Cheers. – Alex Wertheim May 11 '13 at 21:23