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Given that $$ f(x) = \frac{x}{1+x^2} $$

I have to find $$\frac{f(x) - f(a)}{x-a}$$

So some progressing shows that:

$$ \frac{\left(\frac{x}{1+x^2}\right) - \left(\frac{a}{1+a^2}\right)}{x-a} = \frac{(x)(1+a^2)-(a)(1+x^2)}{(1+x^2)(1+a^2)}\cdot\frac{1}{x-a} = \frac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)} $$

Now, is it possible to factor $x+xa^2-a-ax^2$? I can't seem to find a way, as for simplifying the whole thing. Is there any rule I can use, and I'm unable to see?

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Can't you just use the quotient rule to find this derivative? – user2357112 Jun 3 '14 at 21:28
up vote 6 down vote accepted

Since $x-a$ is in the denominator, it makes sense to consider the possibility that $x-a$ is a factor of the numerator. (If you know the factor theorem, you can see this is the case, since the numerator equals zero when $x = a.)$ I see $x-a$ in the numerator, along with two other terms. Very little cleverness is needed at this point to write

$$x-a + xa^2 - ax^2 \; = \; x - a + ax(a-x)$$ $$ = \; (x-a) - ax(x-a) \; = \; (x-a)(1-ax)$$

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Ahhhh. So much time looking at it and I didn't realise this. I don't know about the factor theorem, something to look at. Thank you for your answer. – sidyll Jun 3 '14 at 18:08
@sidyll: For more about using the factor theorem, see my answer at Finding limit of a quotient. For a less obvious use of the factor theorem, see my comments at How to simpify this?. – Dave L. Renfro Jun 3 '14 at 18:12
Thank you for your references – sidyll Jun 3 '14 at 18:17
@sidyll: I just remembered that I posted a couple of old handouts of mine on the factor theorem a couple of years ago. See the files attached to this 12 January 2012 sci.math post archived at Math Forum. – Dave L. Renfro Jun 3 '14 at 18:25
@sidyll As for derivatives, generally the arithmetic works nicely using the quotient rule - see my answer. I'll bet Dave can give links to interesting expositions in AMM, Math. Mag, etc. – Bill Dubuque Jun 3 '14 at 19:02

$$\eqalign{x+xa^2-a-ax^2&= x-a+xa^2-ax^2 \\ & = x-a+x(a^2-ax) \\ &= x-a+x(a(a-x)) \\ &= x-a+x(-a(x-a)) \\ &=\color{blue}{x-a}-ax\color{blue}{(x-a)} \\ &=(x-a)(1-ax).\;\checkmark }$$ Therefore you can conclude that: $$\eqalign{\require{cancel}\dfrac{f(x)-f(a)}{x-a}&=\dfrac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)} \\ &=\dfrac{\color{red}{\cancel{\color{black}{(x-a)}}}(1-ax)}{(1+x^2)(1+a^2)\color{red}{\cancel{\color{black}{(x-a)}}}} \\ &= \dfrac{1-ax}{(1+x^2)(1+a^2)}. }\tag{$x\neq a$}$$

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Thank you very much for your examples in continuing it. – sidyll Jun 3 '14 at 18:09
+1 for nice format. I learn something new about $LaTeX$ from you. – Tunk-Fey Jun 3 '14 at 18:10
@Tunk-Fey I've also learned a lot from your answers and your TeX style! $\overset{\cdot\cdot}\smile$ BTW you can write a more beautiful LaTeX as: $\LaTeX$ using \LaTeX, and there's also a variant for TeX namely $\TeX$ which is produced by \TeX. – Hakim Jun 3 '14 at 18:13
@sidyll You're welcome! Glad I could help! ;-) – Hakim Jun 3 '14 at 18:14
@حكيمالفيلسوفالضائع Thank you for the code. I didn't know that code before. $\ddot\smile$ – Tunk-Fey Jun 3 '14 at 18:17

By the quotient rule for the difference $\ f'(x)\, :=\, \dfrac{f(x)-f(a)}{x-a}$

$$\quad \begin{eqnarray} (g/h)'(x) &=\,\ & \dfrac{\color{#c00}{g'(x)} h(a) - g(a)\color{#0a0}{h'(x)}}{h(a)h(x)}\\ \begin{array}{l}\ g(x)=x\qquad\Rightarrow\,\color{#c00}{g'(x) = 1}\\ h(x) = 1+x^2\,\Rightarrow\,\color{#0a0}{h'(x) = x+a}\\\end{array}\ \Bigg\}\!\!\!\!\!&=& \dfrac{\color{#c00}1\cdot (1+a^2)\overset{\phantom{I^I}}-a(\color{#0a0}{x+a})}{(1+a^2)(1+x^2)}\ =\ \dfrac{1-ax}{(1+a^2)(1+x^2)}\end{eqnarray}$$

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Hint: you can guess that something interesting is going to happen near $x = a$, which suggests looking to factor $x-a$.

Indeed, $x + xa^2 - a - ax^2 = (x-a) + xa(x-a)$.

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Factor: $$x+xa^2−a−ax^2=x-a-ax^2+xa^2=(x-a)-ax(x-a)=(1-ax)(x-a)$$ Thus, $$\frac{f(x)-f(a)}{x-a}=\frac{\frac{(1-ax)(x-a)}{(1+x^2)(1+a^2)}}{(x-a)}=\boxed{\frac{1-ax}{(1+x^2)(1+a^2)}}$$

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