Question 442055
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There are two ways to go about this.  You can use the formula for the derivative of a product:


If *[tex \Large f(x)\ =\ g(x)\,\cdot\,h(x)] then *[tex \Large f'(x)\ =\ g'(x)h(x)\ +\ g(x)h'(x)]


So if you let *[tex \Large g(x)\ =\ 3x\ -\ 2] and *[tex \Large h(x)\ =\ 3x\ +\ 1]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(x)\ =\ 3(3x\ +\ 1)\ +\ 3(3x\ -\ 2)\ =\ 18x\ -\ 3]


Or you could first multiply the two denominators and then take the derivative of the resulting 2nd degree polynomial function.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ 9x^2\ -\ 3x\ -\ 2]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(x)\ =\ 18x\ -\ 3]


Achieving the same result.  Whichever seems easiest to you is the way to do it.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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