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

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2 Answers 2

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.

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Don't forget that differentiable implies continuous. – Hagen von Eitzen Oct 18 '12 at 21:03

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

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