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CALCULUS HELP URGENT!!

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find y' if arctan(xy)=6+x^(4)y

  • CALCULUS HELP URGENT!! - ,

    here is one way:

    arctan(xy)=6+x^(4)y
    or
    xy = tan(6 + x^4y)
    xdy/dx + y = sec^2 (6 + x^4y) (x^4 dy/dx + 4x^3y_
    = x^4 dy/dx sec^2 (6+x^4y) + 4x^3y sec^2(6+x^4y)

    dy/dx(x - x^4sec^2 (6+x^4y) ) = 4x^3 sec^2 (6+x^4y) - y

    dy/dx = (4x^3 sec^2 (6+x^4y) - y)/(x - x^4sec^2 (6+x^4y) )

    check my algebra, I should have written it out on paper first.

  • CALCULUS HELP URGENT!! - ,

    Or, more directly,

    1/(1+x^2y^2) (y+xy') = 4x^3y+x^4y'
    y+xy' = 4x^3y (1+x^2y^2) + x^4(1+x^2y^2)y'
    y' = (4x^3y + 4x^5y^3-y)/(x-x^4-x^6y^2)

    This can also be obtained (after restoring the y that was accidentally dropped) from Reiny's result by recalling the sec^2 = 1+tan^2 = 1+x^2y^2

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