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Hello and thanks in advance for any help!!

I currently have to get to the derivative function of $\frac{a-x}{x}$ by definition.. that is

$$\lim_{h\to0} \frac{\frac{a - (x+h)}{x+h}- \frac{a-x}{x}}{h}$$

So it's kind of a little mess for a newbie in algebra like me. I've tried turning the X into X^-1 with no results, like this:

$$\frac{a(x+h)^{-1} - 1 - ax^{-1} +1}{h}$$

Also, I've tried using a common divisor of $XH(X+H)$ or something, but it's too long to write in $\LaTeX$ (this is my first time using it and I don't feel that it's relevant to the question).

I just wanna know how to begin. That means that I don't want the full answer, just a piece of advice to get me in the right path and then work the exercise out myself.

Thanks a ton everyone!! =)

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    $\begingroup$ Mark, how did you edit the question to render the equation? $\endgroup$ – Gaspa79 Jun 7 '12 at 17:02
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    $\begingroup$ You put $...$ around an inline formula, and $$...$$ around a displayed formula. Also, you can click on the "edited n minutes ago" link to see the changes, as here. $\endgroup$ – MJD Jun 7 '12 at 17:05
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Let $f(x) = \dfrac{a}{x} -1$. Then $$f'(x) = \lim_{h \rightarrow 0} \dfrac{f(x+h) - f(x)}{h} = \lim_{h \rightarrow 0} \dfrac{\left(\dfrac{a}{x+h}-1 \right)- \left( \dfrac{a}{x} - 1\right)}{h}$$ $$f'(x) = \lim_{h \rightarrow 0} \dfrac{ \left(\dfrac{a}{x+h}-1 - \dfrac{a}{x} + 1 \right)}{h} = \lim_{h \rightarrow 0} \dfrac{ \left(\dfrac{a}{x+h} - \dfrac{a}{x}\right)}{h} = \lim_{h \rightarrow 0} \dfrac{ \left(\dfrac{ax - a(x+h)}{x(x+h)}\right)}{h}$$ $$f'(x) = \lim_{h \rightarrow 0} \dfrac{ \left(\dfrac{ax - ax - ah}{x(x+h)}\right)}{h} = \lim_{h \rightarrow 0} \dfrac{ \left(\dfrac{- ah}{x(x+h)}\right)}{h} = \lim_{h \rightarrow 0} \left(\dfrac{- a}{x(x+h)}\right)$$ $$f'(x) = \left(\dfrac{-a}{\displaystyle \lim_{h \rightarrow 0} \left(x(x+h) \right)}\right) = - \dfrac{a}{x \times x} = - \dfrac{a}{x^2}$$

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  • $\begingroup$ How come you never work with align or newline inside $$? $\endgroup$ – Asaf Karagila♦ Jun 7 '12 at 17:27
  • $\begingroup$ @AsafKaragila Sometimes I do. The rest of the times, I find this easier to type than align. $\endgroup$ – user17762 Jun 7 '12 at 17:29
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It's simpler to just put the two fractions in the numerator under a common denominator and then convert the complex fraction into a simple one: $$\begin{align*} \frac{\quad\frac{a-(x+h)}{x+h} - \frac{a-x}{x}\quad}{h} &= \frac{\quad\frac{(a-x-h)x}{(x+h)x} - \frac{(a-x)(x+h)}{x(x+h)}\quad}{h}\\ &=\frac{\quad\frac{(a-x-h)x - (a-x)(x+h)}{x(x+h)}\quad}{h}\\ &= \frac{(a-x-h)x - (a-x)(x+h)}{x(x+h)h}\\ &= \frac{ax-x^2 -xh -ax -ah +x^2 +xh}{xh(x+h)}\\ &= \frac{-ah}{xh(x+h)}\\ &= \frac{-a}{x(x+h)}. \end{align*}$$

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  • $\begingroup$ Thanks a lot man! I would set your answer as accepted too but I can only set one =(. $\endgroup$ – Gaspa79 Jun 7 '12 at 17:55
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If you look only at the fractions in the numerator and combine them, we get $$\frac{(a-x-h)x-(a-x)(x+h)}{x(x+h)}=\frac{(ax-x^2-hx)-(ax+ah-x^2-hx)}{x(x+h)}=\frac{-ah}{x(x+h)},$$ so if we simplify, we are looking for $$\lim_{h\to 0}\frac{a}{x(x+h)}=\frac{-a}{x^2}.$$

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  • $\begingroup$ Wow this is an awesome solution! I would set your answer as accepted too but I can only set one =(. $\endgroup$ – Gaspa79 Jun 7 '12 at 17:56
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    $\begingroup$ Not a problem. I got beat to the punch, after all. ^_^ $\endgroup$ – Cameron Buie Jun 7 '12 at 18:04
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Take common divisor on the numerator and cancel stuff: $$\frac{\frac{a-(x+h)}{x+h}-\frac{a-x}{x}}{h}=\frac{\rlap{//}{ax}-x(\rlap{/}{x}+\rlap{/}{h})-\rlap{//}{ax}-a\rlap{/}{h}+\rlap{//}{x^2}+\rlap{//}{xh}}{x\rlap{/}{h}(x+h)}\to-\frac{a}{x^2}$$

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  • $\begingroup$ I can't believe you worked it out in just one line. I would set your answer as accepted too but I can only set one =(. I upvoted of course =) $\endgroup$ – Gaspa79 Jun 7 '12 at 17:56
  • $\begingroup$ Don't worry about it. It helps a lot in higher mathematics to sharpen high school algebra skills: factorisation, common divisor/multiple, binomial/trinomial expressions, etc. $\endgroup$ – DonAntonio Jun 7 '12 at 21:13
  • $\begingroup$ I will certainly do so. Thanks a lot again. I'm guessing that since your name is Antonio, you speak spanish, do you? Muchas gracias desde Argentina che! Voy a practicar comun divisor que parece bastante util =). $\endgroup$ – Gaspa79 Jun 7 '12 at 21:30
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    $\begingroup$ @Gaspa Todo lo que viste en el secundario de álgebra: reducciones de fracciones, con o sin exponentes, fracciones mismas, las fórmulas para $$(a\pm b)^2\,\,,\,a^2-b^2$$ etc. Saludos desde México. $\endgroup$ – DonAntonio Jun 7 '12 at 21:34

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