Posted by sara on Wednesday, October 31, 2007 at 10:34pm.
The derivatives of these trig functions should be right there in your Calculus text.
for the second of your questions, I will use y' for the first derivative and y" for the second derivative i.e. y'= dy/dx
1+y=x+xy so
y' = 1 + y + xy'
y'(1-x) = 1+y
y' = (1+y)/(1-x)
y" = [(1-x)y' - (1+y)(-1)]/((1-x)^2)
=[(1+y) + (1+y)]/((1-x)^2) after I replaced y'
=(2+2y)/((1-x)^2)
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