# Derivative using limit definition

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

• Welcome to Math.SE. It would help if you showed us what you have tried, so we know how to help. – Empy2 Nov 2 '13 at 5:28
• $g'(x)=\lim_{h\to 0} \frac{g(x+h)-g(x)}{h}= \lim_{h\to 0} \frac{(x+h)f(x+h)-xf(x)}{h} = \frac{(x+h)f(x+h)-xf(x)}{h}= x(f(x+h)-f(x)+h(f(x+h)-f(x)) =$ – Worawit Tepsan Nov 2 '13 at 5:37

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 $\lim_{h\to 0}f(x+h)=f(x)$.
• 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)? – auraham Nov 2 '13 at 5:53
• 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. – André Nicolas Nov 2 '13 at 5:59
$\frac{g(x+h)-g(x)}{h} = \frac{(x+h)f(x+h)- xf(x)}{h} = x\frac{f(x+h)- f(x)}{h}+ f(x+h)$.
Hence $\lim_{h \to 0} \frac{g(x+h)-g(x)}{h} = \lim_{h \to 0} \left( x\frac{f(x+h)- f(x)}{h}+ f(x+h) \right) = xf'(x)+f(x)$.