Factoring a derivative

jmoney30

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I'm reading a book calculus made easy because I love math , it keeps my mine fresh and I can help my kids with homework. In the book, he's factoring but not sure where he's getting these numbers. Please see below. I'm not sure where the 3/2 and 1/2 are coming from when it appears some of the values should cancel out. Please help or correct me .

Thanks

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I'm not sure where the 3/2 and 1/2 are coming from
Hi jmoney. If we distribute the 1/2 in the line above their result, then we can remove all grouping symbols. There will be two terms with exponent 1/2, two terms with exponent -1/2, two terms with exponent -3/2 and two terms with exponent -5/2. Combine each pair of like-terms, yielding a total of four terms. Two of them will be multiplied by 3/2, and two will be multiplied by 1/2..

EG:
[imath]\;\; \frac{1}{2}\theta^{½} + \theta^{½} \;=\; \frac{1}{2}\theta^{½} + \frac{2}{2}\theta^{½} \;=\; \frac{1+2}{2}\theta^{½} \;=\; \frac{3}{2}\theta^{½} \;\;[/imath] :)
[imath]\;[/imath]
 
This just restates, in mathematical notation, what Otis already said.

[math] \frac{1}{2} * ( \theta ^{1/2} + \theta^{-3/2} - \theta^{-1/2} - \theta^{-5/2}) + ( \theta^{1/2} + \theta^{-1/2} - \theta^{-3/2} - \theta^{-5/2} ) = \\ \frac{1}{2} * ( \theta ^{1/2} + \theta^{-3/2} - \theta^{-1/2} - \theta^{-5/2}) + \frac{1}{2} * (2 \theta^{1/2} + 2 \theta^{-1/2} - 2 \theta^{-3/2} - 2 \theta^{-5/2} ) = \\ \frac{1}{2} * (3 \theta ^{1/2} - \theta^{-3/2} + \theta^{-1/2} - 3 \theta^{-5/2}) = \\ \frac{3}{2} * ( \theta^{1/2} - \theta^{-5/2}) + \frac{1}{2} * ( \theta^{1/2} - \theta^{-3/2}). [/math]
Lot of fussy algebra was skipped over,
 
Thanks I see , I was trying to add the two together before putting a common denominator of 2 in the second term. This is why my terms keep canceling out. I should’ve saw this.Thanks
 
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