Question 1034339
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(x)\ =\ (6x\ -\ 2)\sqrt{x^2\ -\ 5x\ +\ 3}]


Let *[tex \Large u\ =\ 6x\ -\ 2]


Let *[tex \Large w =\ x^2\ -\ 5x\ +\ 3]


Let *[tex \Large v\ =\ \sqrt{w}]


Then using the product rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ v\frac{du}{dx}\ +\ u\frac{dv}{dx}]


But first we need to use the chain rule to find *[tex \Large\ \frac{dv}{dx}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dv}{dx}\ =\ \frac{dv}{dw}\cdot\frac{dw}{dx}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dv}{dw}\ =\ \frac{1}{2\sqrt{w}}\ =\ \frac{1}{2\sqrt{x^2\ -\ 5x\ +\ 3}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dw}{dx}\ =\ \2x\ -\ 5]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dv}{dx}\ = \frac{2x\ -\ 5}{2\sqrt{x^2\ -\ 5x\ +\ 3}}]


Now back to the product rule, noting that *[tex \Large \frac{du}{dx}\ =\ 6]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ v\frac{du}{dx}\ +\ u\frac{dv}{dx}]


We just substitute all of the parts that we now know:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ 6\sqrt{x^2\ -\ 5x\ +\ 3}\ +\ \frac{(6x\ -\ 2)(2x\ -\ 5)}{2\sqrt{x^2\ -\ 5x\ +\ 3}}]


This result demands simplification, but I'll leave the rest in your capable hands.


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 \ \ \ \ \ \ \ \ \ \  

</font>