Posted by abhi on Monday, December 24, 2012 at 3:19pm.
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
Should read:
y"=0+2sec(x)*(-1/cosē(x))(-sin(x))
=2secē(x)tan(x)
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