Calculus

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Find the equation of the tangent line to the graph of √(xy)=x-2y at (4,1)

  • Calculus -

    (1/2)(xy)^(-1/2) (x dy/dx + y) = 1 - 2dy/dx
    at the given point

    (1/2)(4)^(-1/2) (4dy/dx + 1) = 1 - 2dy/dx
    (1/4)(4dy/dx + 1 = 1 - 2dy/dx
    times 4
    4dy/dx + 4 = 1 - 2dy/dx
    6dy/dx = -3
    dy/dx = -1/2

    y-1 = -(1/2)(x-4)
    2y-2= -x+4
    x + 2y = 6
    or
    y = (-1/2)x + 6

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