# Factoring exponents that differ by an integer

• Feb 6th 2013, 05:00 PM
camjerlams
Factoring exponents that differ by an integer
Attachment 26887

Hi guys sending this on my phone at work please tell me if its too small and ill type it out. This is the full working that I have for a derivative problem which I was fine with until the very last step. When I did it my last step involved putting the first term back into a square root and the same with the last term except in the denominator under 1.
Some factoring has happened here that I don't understand, can someone please explain to me what has happened and why between the last 2 steps? (The denominator at bottom has been cut out on uploading, it's a 3)
• Feb 6th 2013, 05:54 PM
HallsofIvy
Re: Factoring exponents that differ by an integer
The "last two steps" are just combining like terms. The derivative of $x(x+ 3)^{1/2}$ is, as you say, $(x+ 3)^{1/2}+ (1/2)x(x+ 3)^{-1/2}$. You can, then, factor $(x+ 3)^{-1/2}$. Of course, $(x+ 3)^{1/2}= (x+ 3)(x+ 3)^{-1/2}$ so that gives $(x+ 3)^{-1/2}(x+ 3- (1/2)x)$. That denominator that was cut off should be a 2, not a 3. And x- (1/2)x= (1/2)x so that is equal to $(x+ 3)^{-1/2}((1/2)x+ 3)$