Monday

September 1, 2014

September 1, 2014

Posted by **Anna** on Sunday, October 3, 2010 at 12:11am.

I know that I would start with

f'(x)=2((x^2+3)^5+x) but I don't know how to do the rest

- Calculus -
**Anna**, Sunday, October 3, 2010 at 12:14amThe answer is

f'(x)=2((x^2+3)^5+x)[5(x^2+3)^4(2x)+1]

I don't understand where the +1 at the end comes from

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