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

If y=x+tanx then prove that cos^2x(y2)-2y+2x=0 where y2 is the second derivative...plz help me out.

- maths -
**MathMate**, Monday, December 24, 2012 at 9:32pm
y=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 9:33pm
Should read:

y"=0+2sec(x)*(-1/cos²(x))(-sin(x))

=**2**sec²(x)tan(x)

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