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Calculus

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y=arctan(sqrt(9x^2-1))^2
dy/dx=?

  • Calculus - ,

    since (√(9x^2-1))^2 = 9x^2-1,
    I assume you mean arctan^2 √(9x^2-1)
    let u = 9x^2-1
    y = arctan^2 (√u)
    y' = 2 arctan(√u) * 1/(u+1) * 1/(2√u) u'
    = 2 arctan(√9x^2-1) * 1/(9x^2) * 1/(2√(9x^2-1)) * 18x
    = 2 arctan(√9x^2-1) / [x √(9x^2-1)]

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