1
$\begingroup$

Using first principle, prove that if$$g(x) = x\cdot f(x)$$ then $$g'(x)= x \cdot f'(x) + f(x)$$

I tried this:

$$g'(x) = \lim_{h\to 0} \frac{[(x+h)\cdot f(x+h) - (x\cdot f(x))]}{h}$$

$\endgroup$
  • $\begingroup$ Can you elaborate on what you mean by "first principle"? Where did you use it in your attempt? $\endgroup$ – Thibaut Dumont Mar 7 '17 at 22:36
  • 1
    $\begingroup$ First principle is $f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$ $\endgroup$ – lakada Mar 7 '17 at 22:36
  • 3
    $\begingroup$ so far so good. Now collect the terms $\frac {x(f(x+h) - f(x)) + h(f(x+h)}{h}$ and evaluate the limit $\endgroup$ – Doug M Mar 7 '17 at 22:36
5
$\begingroup$

You almost got it! Just readjust the terms like that

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

$\endgroup$
  • $\begingroup$ Nice answer and way simpler and more direct than mine. +1 $\endgroup$ – DonAntonio Mar 7 '17 at 23:00
  • $\begingroup$ $g'(x)=\lim_{h\to 0} \frac{[\color{green}{(x+h)f(x+h)}-xf(x)]}{h} = x\lim_{h\to 0}\frac{\color{green}{f(x+h)}-f(x)}{h}+\lim_{h\to 0}\frac{hf(x+h)}{h}$ How does $(x+h)$ separate from $f(x+h)$ $\endgroup$ – lakada Mar 7 '17 at 23:17
  • 1
    $\begingroup$ @lakada $(x+h)\cdot f(x+h) = x \cdot f(x+h) + h \cdot f(x+h)$ $\endgroup$ – dxiv Mar 8 '17 at 0:10
  • $\begingroup$ @dvix thanks makes sense now $\endgroup$ – lakada Mar 8 '17 at 0:21
  • 1
    $\begingroup$ $g'(x) = \lim_{h\to 0} \frac{g(x+h)- g(x)}{h} =\lim_{h\to 0} \frac{[(x+h)f(x+h)-xf(x)]}{h} = \lim_{h\to 0} \frac{[x\cdot f(x+h) + h\cdot f(x+h)-xf(x)]}{h} = \lim_{h\to 0} \frac{[x\cdot f(x+h)-xf(x)]}{h} + \frac{h\cdot f(x+h)}{h} = \lim_{h\to 0} \frac{x\cdot[ f(x+h)-f(x)]}{h} + \lim_{h\to 0} \frac{h\cdot f(x+h)}{h} = x \cdot f'(x) + f(x)$ $\endgroup$ – lakada Mar 8 '17 at 0:43
2
$\begingroup$

$$\lim_{h\to0}\frac{g(x+h)-g(x)}h=\lim_{h\to0}\frac{(x+h)f(x+h)-xf(x)}h=$$

$$=\lim_{h\to0}\frac{(\color{green}x+\color{red}h)f(x+h)-\color{red}{hf(x)}+hf(x)-\color{green}{xf(x)}}h=$$

$$=\lim_{h\to0}\frac{\color{red}{h\left[f(x+h)-f(x)\right]}+\color{green}{x\left[f(x+h)-f(x)\right]}}h+\overbrace{f(x)}^{=\frac{hf(x)}h}=$$

$$=\lim_{h\to0}\require{cancel}\frac{\cancel h\left[f(x+h)-f(x)\right]}{\cancel h}+\lim_{h\to0}x\frac{\left[f(x+h)-f(x)\right]}h+f(x)=$$

$$=0+xf'(x)+f(x)=xf'(x)+f(x)$$

Observe that first zero is due to continuity of $\;f\;$ at each $\;x\;$ where it is differentiable

$\endgroup$
  • $\begingroup$ how do you get $hf(x)$ in the 2nd line? $\endgroup$ – lakada Mar 7 '17 at 22:43
  • $\begingroup$ @lakada Old trick: I added and substracted $\;hf(x)\;$ in the numerator. This works in several cases. $\endgroup$ – DonAntonio Mar 7 '17 at 22:44
  • $\begingroup$ In the 3rd line how did you create those factors? $\endgroup$ – lakada Mar 7 '17 at 22:49
  • $\begingroup$ @lakada Grouping from the second line...Perhaps I'll add some colours to make it clearer. $\endgroup$ – DonAntonio Mar 7 '17 at 22:50
  • $\begingroup$ Is there another way to solve this question without grouping? $\endgroup$ – lakada Mar 7 '17 at 22:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.