calculus

1. Given the curve
a. Find an expression for the slope of the curve at any point (x, y) on the curve.
b. Write an equation for the line tangent to the curve at the point (2, 1)
c. Find the coordinates of all other points on this curve with slope equal to the slope at (2, 1)

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  1. sorry here's the equation:
    x+xy+2y^2=6

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  2. x+xy+2y^2=6

    (a)
    x(1+y)=6-2y²
    x=(6-2y²)/(1+y)
    differentiate with respect to y to get
    dx/dy=-4x/(x+1)-(6-2x^2)/(x+1)^2
    So
    dy/dx
    = 1/(dx/dy)
    = -(1/2)(x^2+2x+1)/(x^2+2x+3)

    (b) For the point (2,1), first check that it lies on the curve.
    f'(2)=-(1/2)(2^2+2*2+1)/(2^2+2*2+3)
    =-9/22

    (c)
    Now solve for
    f'(x)=-9/22
    to get x=-4 or x=2

    Verify all the numerical values.

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  3. There was a mistake in the calculation of dx/dy. It should have been expressed in terms of y and not x.

    Given : x+xy+2y^2=6
    (a)
    x(1+y)=6-2y²
    x=(6-2y²)/(1+y)
    differentiate with respect to y to get
    dx/dy=-4y/(y+1)-(6-2y^2)/(y+1)^2
    So
    f'(y)=dy/dx
    = 1/(dx/dy)
    = -(1/2)(y^2+2y+1)/(y^2+2y+3)

    (b) For the point (2,1), first check that it lies on the curve.

    f'(y)=
    f'(1)=-(1/2)(1^2+2*1+1)/(1^2+2*1+3)
    =-1/3

    Tangent line with slope -1/3 passing through (2,1) is:
    (y-1) = (-1/3)(x-2)

    (c)
    Now solve for
    f'(y)=-1/3
    to get x=-3 or x=1

    Please verify all the numerical values.

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