Question 1112736
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One problem per post.


5.  *[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{d}{dx}\ \[\frac{\(x^2\ +\ 5\)^2}{\(x^2\ -\ 5\)^2}\] ]


Let *[tex \Large u\ =\ x^2\ +\ 5] and let *[tex \Large v\ =\ x^2\ -\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{d}{dx}\ \[\frac{u^2}{v^2}\]\ =\ \frac{\frac{d}{dx}u^2\cdot v^2\ -\ u^2\cdot \frac{d}{dx}v^2}{v^4}]


By the quotient rule


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =\ \frac{2u\frac{du}{dx}\cdot v^2\ -\ 2v\frac{dv}{dx}\cdot u^2}{v^4}]


By the chain rule


But *[tex \Large \frac{du}{dx}\ =\ \frac{dv}{dx}\ =\ 2x]


So substituting back:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4x\(x^2\ +\ 5\)\(x^2\ -\ 5\)^2\ -\ 4x\(x^2\ -\ 5\)\(x^2\ +\ 5\)^2}{\(x^2\ -\ 5\)^4]


A little more algebra...


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{d}{dx}\ \[\frac{\(x^2\ +\ 5\)^2}{\(x^2\ -\ 5\)^2}\]\ =\ -\frac{40\(x^2\ +\ 5\)}{\(x^\ -\ 5\)^3}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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