# A curve has the equation (x+y)^2 = xy^2. Find the gradient of the curve at the point where x=1

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The first step is to find dy/dx.

To do this you must first expand the brackets.

x2 + y2 + 2xy = xy2

Then differentiate each term with respect to x.

dy/dx of (x2) = 2x

dy/dx of (y2) = 2y(dy/dx)

(Using the product rule with u=yand v=1) this can be explained in more detail if necessary.

dy/dx of (2xy) = 2y + 2x(dy/dx)

(Also using the product rule with u=2x and v=y)

dy/dx of (xy2) = y2 + 2xy(dy/dx)

(Also product rule with u=x and v=y2)

Overall that gives:

2x + 2y + 2x(dy/dx) + 2y(dy/dx) = y2 + 2xy(dy/dx)

Then put all the terms containing (dy/dx) on the left and the others on the right.

This gives:

2x(dy/dx) + 2y(dy/dx) - 2xy(dy/dx) = y- 2x - 2y

This equals:

(dy/dx)(2x + 2y - 2xy) = y2 - 2x - 2y

Therefore:

(dy/dx) = (y2 - 2x - 2y) / (2x + 2y - 2xy)

From the original equation you now need to work at y at the point when x=1.

(1+y)2 = 1y2

y2 + 2y + 1 = y2

This is the same as:

2y + 1 = 0

2y = -1

y= -0.5

Substitute this into (dy/dx)

This gives:

[(-0.5)2 - 2(1) -2(-0.5)] / [2(1) + 2(-0.5) - 2(1 x -0.5)]

This equals

(0.25 - 2 + 1) / (2 - 1 +1)

Therefore the gradient equals -3/8

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