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)2x2tan(x)+2x
=0

maths  MathMate, Monday, December 24, 2012 at 9:33pm
Should read:
y"=0+2sec(x)*(1/cos²(x))(sin(x))
=2sec²(x)tan(x)
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