Successive Differentiation and division

I am working on successive differentiation. I have ran into some confusion and would like some help with the process of differentiating when dealing with division. Here is the problem that sparked my intentions to post here: $$y = \frac{x^2 + a}{x + a}$$ I am reading Calculus Made Easy by Silvanus P. Thompson and unfortunately there is no example on how I should go about this. I can find $\frac{dy}{dx}$ but can't seem to find $\frac{d^2 y}{dx^2}$ for this exercise. I find $\frac{dy}{dx}$ by doing this: $$y = \frac{(x + a)(2x) - (x^2 + a)(1)}{(x + a)^2}$$ to get $y = \frac{x^2 + 2xa - a}{(x + a)^2}$. This is where I am stuck. I have attempted to simply perform the same operation on this result to differentiate again but it isn't adding up. Any help on this is much appreciated!

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Small edit: you sure you don't mean $\frac{dy}{dx}$? ;) – Kendra Lynne Jul 15 '13 at 19:23
Many thanks Kendra :) – Blake_B Jul 15 '13 at 19:27
In this case you can simplify $x^2 + 2xa - a = (x+a)^2 - a(a+1)$, which makes the further derivatives nice. In general, though, yes differentiation of quotients quickly leads to big unwieldy expressions. – Daniel Fischer Jul 15 '13 at 19:29
When using the quotient rule on $f\over g$, it's often better to calculate and simplify $f'$ and $g'$ first, then calculate and simplify $f'g$ and $fg'$, and finally piece it all together. – Jack M Jul 15 '13 at 20:00

$$\dfrac{dy}{dx} = \frac{(x + a)(2x) - (x^2 + a)(1)}{(x + a)^2} =\dfrac{2x^2 + 2ax - x^2 - a}{(x+a)^2} = \frac{x^2 + 2xa - a}{(x + a)^2}$$ So far so good.

Now we need to differentiate again, using the quotient rule, as you did when finding $\dfrac{dy}{dx}$. It gets messy-looking, but in this case, simplifies relatively nicely in the end:

$$\dfrac{d^2y}{dx^2} = \dfrac{(x+a)^2(2x + 2a) - (x^2 + 2xa - a)[2(x+a)(1)]}{((x+ a)^2)^2}$$

We can factor the numerator, expand, and then simplify:

\begin{align}\dfrac{d^2y}{dx^2} & = \dfrac{2(x+a)[(x+a)^2 - (x^2 + 2xa - a)]}{(x+ a)^4} \\ \\ & = \dfrac{2[x^2 + 2ax + a^2 - x^2 - 2ax + a]}{(x+ a)^3}\\ \\ & =\dfrac{2( a^2 + a)}{(x + a)^3} \end{align}

We could have gotten clever by rewriting $\dfrac{dy}{dx}$: We can rewrite $\dfrac{dy}{dx}$ thusly:

\begin{align} \dfrac{dy}{dx} = \frac{x^2 + 2xa - a}{(x + a)^2} & = \dfrac{x^2 + 2xa + a^2 -a -a^2}{(x+ a)^2} \\ \\ & = \dfrac{(x+a)^2 - a(a + 1)}{(x + a)^2} \\ \\ & = 1 - \dfrac{a(a + 1)}{(x + a)^2}\end{align}

The result of differentiating the result (using essentially, the power rule) would be the same as given above.

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Amzoti: Yes it did, but you've got me beat when it comes to patience time-involved-to-format! ;-) – amWhy Jul 16 '13 at 1:37