Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!! =)

share|improve this question
1  
Mark, how did you edit the question to render the equation? –  Damieh Jun 7 '12 at 17:02
2  
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. –  MJD Jun 7 '12 at 17:05

4 Answers 4

up vote 5 down vote accepted

Let $f(x) = \dfrac{a}{x} -1$. Then $$f'(x) = \lim_{h \rightarrow 0} \dfrac{f(x+h) - f(x)}{h} = \lim_{h \rightarrow 0} \dfrac{\left(\dfrac{a}{x+h}-1 \right)- \left( \dfrac{a}{x} - 1\right)}{h}$$ $$f'(x) = \lim_{h \rightarrow 0} \dfrac{ \left(\dfrac{a}{x+h}-1 - \dfrac{a}{x} + 1 \right)}{h} = \lim_{h \rightarrow 0} \dfrac{ \left(\dfrac{a}{x+h} - \dfrac{a}{x}\right)}{h} = \lim_{h \rightarrow 0} \dfrac{ \left(\dfrac{ax - a(x+h)}{x(x+h)}\right)}{h}$$ $$f'(x) = \lim_{h \rightarrow 0} \dfrac{ \left(\dfrac{ax - ax - ah}{x(x+h)}\right)}{h} = \lim_{h \rightarrow 0} \dfrac{ \left(\dfrac{- ah}{x(x+h)}\right)}{h} = \lim_{h \rightarrow 0} \left(\dfrac{- a}{x(x+h)}\right)$$ $$f'(x) = \left(\dfrac{-a}{\displaystyle \lim_{h \rightarrow 0} \left(x(x+h) \right)}\right) = - \dfrac{a}{x \times x} = - \dfrac{a}{x^2}$$

share|improve this answer
    
How come you never work with align or newline inside $$? –  Asaf Karagila Jun 7 '12 at 17:27
    
@AsafKaragila Sometimes I do. The rest of the times, I find this easier to type than align. –  user17762 Jun 7 '12 at 17:29

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: $$\begin{align*} \frac{\quad\frac{a-(x+h)}{x+h} - \frac{a-x}{x}\quad}{h} &= \frac{\quad\frac{(a-x-h)x}{(x+h)x} - \frac{(a-x)(x+h)}{x(x+h)}\quad}{h}\\ &=\frac{\quad\frac{(a-x-h)x - (a-x)(x+h)}{x(x+h)}\quad}{h}\\ &= \frac{(a-x-h)x - (a-x)(x+h)}{x(x+h)h}\\ &= \frac{ax-x^2 -xh -ax -ah +x^2 +xh}{xh(x+h)}\\ &= \frac{-ah}{xh(x+h)}\\ &= \frac{-a}{x(x+h)}. \end{align*}$$

share|improve this answer
    
Thanks a lot man! I would set your answer as accepted too but I can only set one =(. –  Damieh Jun 7 '12 at 17:55

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 $$\lim_{h\to 0}\frac{a}{x(x+h)}=\frac{-a}{x^2}.$$

share|improve this answer
    
Wow this is an awesome solution! I would set your answer as accepted too but I can only set one =(. –  Damieh Jun 7 '12 at 17:56
1  
Not a problem. I got beat to the punch, after all. ^_^ –  Cameron Buie Jun 7 '12 at 18:04

Take common divisor on the numerator and cancel stuff: $$\frac{\frac{a-(x+h)}{x+h}-\frac{a-x}{x}}{h}=\frac{\rlap{//}{ax}-x(\rlap{/}{x}+\rlap{/}{h})-\rlap{//}{ax}-a\rlap{/}{h}+\rlap{//}{x^2}+\rlap{//}{xh}}{x\rlap{/}{h}(x+h)}\to-\frac{a}{x^2}$$

share|improve this answer
    
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 =) –  Damieh Jun 7 '12 at 17:56
    
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. –  DonAntonio Jun 7 '12 at 21:13
    
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 =). –  Damieh Jun 7 '12 at 21:30
1  
@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. –  DonAntonio Jun 7 '12 at 21:34

Your Answer

 
discard

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.