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March 6, 2015

March 6, 2015

Posted by **sara** on Wednesday, October 31, 2007 at 10:34pm.

y=2cotx+sec3x+cscX

find d^2/dx^2 for 1+y=x+xy

- derivatives -
**Reiny**, Wednesday, October 31, 2007 at 10:54pmThe 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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