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December 18, 2014

December 18, 2014

Posted by **abhi** on Monday, December 24, 2012 at 3:19pm.

- maths -
**MathMate**, Monday, December 24, 2012 at 9:32pmy=x+tan(x)

y'=1+sec(x)^2

y"=0+2sec(x)*(-1/cos²(x))(-sin(x))

=sec²(x)tan(x)

=>

cos²(x)y"-2y+2x

=2cos²(x)sec²(x)tan(x)-2(x+tan(x)+2x

=2tan(x)-2x-2tan(x)+2x

=0

- maths -
**MathMate**, Monday, December 24, 2012 at 9:33pmShould read:

y"=0+2sec(x)*(-1/cos˛(x))(-sin(x))

=**2**sec˛(x)tan(x)

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