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

  • maths -

    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 -

    Should read:

    y"=0+2sec(x)*(-1/cos²(x))(-sin(x))
    =2sec²(x)tan(x)

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