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Given that f is a differentiable function and g(x) = xf(x). Use the limit definition of derivative show that g'(x) = xf'(x) + f(x).

I understand that you have to find the derivative of xf(x) using the difference quotient but when I set up the problem I can't really simplify it.

Any ideas?

This is what I have so far:

Lim as h -> 0

(x+h) f(x+h) - xf(x) / h

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    $\begingroup$ Welcome to Math.SE. It would help if you showed us what you have tried, so we know how to help. $\endgroup$ – Empy2 Nov 2 '13 at 5:28
  • $\begingroup$ $g'(x)=\lim_{h\to 0} \frac{g(x+h)-g(x)}{h}= \lim_{h\to 0} \frac{(x+h)f(x+h)-xf(x)}{h} = \frac{(x+h)f(x+h)-xf(x)}{h}= x(f(x+h)-f(x)+h(f(x+h)-f(x)) =$ $\endgroup$ – Worawit Tepsan Nov 2 '13 at 5:37
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Let $g(x)=xf(x)$. we want to calculate $$\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}.$$ Substituting, we find that we want $$\lim_{h\to 0} \frac{(x+h)f(x+h)-xf(x)}{h}.$$ Expand. We want $$\lim_{h\to 0} \left(\frac{xf(x+h)-xf(x)}{h}+f(x+h)\right).$$ This is $$x\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}+\lim_{h\to 0} f(x+h).$$ To finish, note that since $f$ is differentiable at $x$, it is continuous at $x$, so $\lim_{h\to 0}f(x+h)=f(x)$.

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  • $\begingroup$ Okay, now it makes sense. Thanks! $\endgroup$ – user104827 Nov 2 '13 at 5:43
  • $\begingroup$ You are welcome. The proofs of most of the "rules of differentiation" use similar ideas. $\endgroup$ – André Nicolas Nov 2 '13 at 5:44
  • $\begingroup$ I don't understand what do you mean on "note that since $f$ is differentiable at $x$ ... so ...", how the limit $x lim_{h -> 0} ... $ (forth equation) becomes to $lim_{h->0}$ (last equation)? $\endgroup$ – auraham Nov 2 '13 at 5:53
  • $\begingroup$ No, you should answer questions and question answers :-). $\endgroup$ – copper.hat Nov 2 '13 at 5:59
  • $\begingroup$ In the last display, we have $x\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$, which I left to you to recognize as $xf'(x)$. Then $\lim_{h\to 0}f(x+h)$ remains to be evaluated. Thus our answer will be $xf'(x)+f(x)$. I can prove the $\lim_{h\to 0}f(x+h)$ part if you wish. Probably this has been done in your course. $\endgroup$ – André Nicolas Nov 2 '13 at 5:59
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$\frac{g(x+h)-g(x)}{h} = \frac{(x+h)f(x+h)- xf(x)}{h} = x\frac{f(x+h)- f(x)}{h}+ f(x+h)$.

Hence $\lim_{h \to 0} \frac{g(x+h)-g(x)}{h} = \lim_{h \to 0} \left( x\frac{f(x+h)- f(x)}{h}+ f(x+h) \right) = xf'(x)+f(x)$.

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