Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $f$ is a differentiable function and $g(x)=xf(x)$, use the definition of a derivative to show that $g'(x)=xf'(x)+f(x)$.

share|cite|improve this question

Just do it: set up the difference quotient and take the limit as $h\to 0$. I’ll get you started:

$$\begin{align*} g'(x)&=\lim_{h\to 0}\frac{g(x+h)-g(x)}h\\ &=\lim_{h\to 0}\frac{(x+h)f(x+h)-xf(x)}h\\ &=\lim_{h\to 0}\frac{x\big(f(x+h)-f(x)\big)+hf(x+h)}h\\ &=\lim_{h\to 0}\frac{x\big(f(x+h)-f(x)\big)}h+\lim_{h\to 0}\frac{hf(x+h)}h\;. \end{align*}$$

Now just finish working out what those last two limits are; it shouldn’t be hard, especially when you already know what they must be.

share|cite|improve this answer
Don't forget that differentiable implies continuous. – Hagen von Eitzen Oct 18 '12 at 21:03

By definition $$ g'(x) = \lim_{\Delta x \to 0} \frac{g(x + \Delta x) - g(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{(x + \Delta x)f(x + \Delta x) - xf(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{x[f(x + \Delta x) - f(x)] + \Delta xf(x + \Delta x)}{\Delta x} = xf'(x) + f(x)$$

share|cite|improve this answer

Your Answer


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.