Posted by Jes on Tuesday, October 30, 2012 at 12:42am.
#1. (e^x)' = e^x, so
f' = 2e^x - 2x
#2.
quotient rule:
f' = (e^x(e^x+1) - e^x*e^x)/(e^x+1)^2
= (e^2x + e^x - e^2x)/(e^x+1)^2
= e^x/(e^x+1)^2
or, knowing that e^x/(e^x+1) = 1 - 1/(e^x+1)
f' = e^x/(e^x+1)^2
#3. Recall that (a^u)' = ln a a^u u'
y' = ln 10 10^√x * 1/(2√x) = ln10/(2√x) 10^(√x)
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