Prove the derivative of $(a-x) / x$ by definition 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!! =)
 A: Let $f(x) = \dfrac{a}{x} -1$. Then $$f'(x) = \lim_{h \rightarrow 0} \dfrac{f(x+h) - f(x)}{h} = \lim_{h \rightarrow 0} \dfrac{\left(\dfrac{a}{x+h}-1 \right)- \left( \dfrac{a}{x} - 1\right)}{h}$$
$$f'(x) = \lim_{h \rightarrow 0} \dfrac{ \left(\dfrac{a}{x+h}-1 - \dfrac{a}{x} + 1 \right)}{h} = \lim_{h \rightarrow 0} \dfrac{ \left(\dfrac{a}{x+h} - \dfrac{a}{x}\right)}{h} = \lim_{h \rightarrow 0} \dfrac{ \left(\dfrac{ax - a(x+h)}{x(x+h)}\right)}{h}$$
$$f'(x) = \lim_{h \rightarrow 0} \dfrac{ \left(\dfrac{ax - ax - ah}{x(x+h)}\right)}{h} = \lim_{h \rightarrow 0} \dfrac{ \left(\dfrac{- ah}{x(x+h)}\right)}{h} = \lim_{h \rightarrow 0} \left(\dfrac{- a}{x(x+h)}\right)$$
$$f'(x) = \left(\dfrac{-a}{\displaystyle \lim_{h \rightarrow 0} \left(x(x+h) \right)}\right) = - \dfrac{a}{x \times x} = - \dfrac{a}{x^2}$$
A: 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:
$$\begin{align*}
\frac{\quad\frac{a-(x+h)}{x+h} - \frac{a-x}{x}\quad}{h} &= 
\frac{\quad\frac{(a-x-h)x}{(x+h)x} - \frac{(a-x)(x+h)}{x(x+h)}\quad}{h}\\
&=\frac{\quad\frac{(a-x-h)x - (a-x)(x+h)}{x(x+h)}\quad}{h}\\
&= \frac{(a-x-h)x - (a-x)(x+h)}{x(x+h)h}\\
&= \frac{ax-x^2 -xh -ax -ah +x^2 +xh}{xh(x+h)}\\
&= \frac{-ah}{xh(x+h)}\\
&= \frac{-a}{x(x+h)}.
\end{align*}$$
A: 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 $$\lim_{h\to 0}\frac{a}{x(x+h)}=\frac{-a}{x^2}.$$
A: Take common divisor on the numerator and cancel stuff: $$\frac{\frac{a-(x+h)}{x+h}-\frac{a-x}{x}}{h}=\frac{\rlap{//}{ax}-x(\rlap{/}{x}+\rlap{/}{h})-\rlap{//}{ax}-a\rlap{/}{h}+\rlap{//}{x^2}+\rlap{//}{xh}}{x\rlap{/}{h}(x+h)}\to-\frac{a}{x^2}$$
