Calculus

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what is the equation of the tangent line that passes through

X^(1/4) + y^(1/4) =4 at the point (16,16)

please give the equation

  • Calculus -

    (1/4)x^(-3/4) + (1/4)y^(-3/4) dy/dx = 0
    dy/dx = -x^(-3/4)/y^(-3/4)
    = y^(3/4)/x^(3/4)

    sub in the point(16,16) to get the slope.
    since you have a point on the line and the slope, it becomes an easy question.

    btw, 16^(3/4) = [ 16^(1/4) ]^3 = 2^3 = 8
    so slope = 8/8 = 1

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