# Find the slope of the curve x^2 + xy + y^2 = 7 at (1,2)

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### 3 Answers

x^2 + xy + y^2 = 7

The slope is the derivative at (1,2).

Let us diffirentiate with respect to x:

=(x^2 + xy + y^2)' = (7)'

=(x^2)' + (xy)' + (y^2)' = (7)'

= 2x + (y' + y) + 2yy' = 0

==> 2x + y' + y + 2yy' = 0

==> Now substitue with (1,2)

==> 2*1 + y' + 2 + 2*2*y' = 0

==> 2 + y' + 2 + 4y' = 0

==> 4 + 5y' = 0

==> 5y' = -4

==> y' = -4/5

**Then the slope is -4/5**

To do this problem we first apply the product rule:

d ( xy) / dx = y + x*dy/dx

to find the derivative of x^2 + xy + y^2 = 7.

So the derivative of x^2 + xy + y^2 = 7 is

2x + x*dy/dx +y + 2y* dy/dx =0

Now taking dy/dx to one side,

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

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

At (1,2)

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

=>( - 1^2 - 1*2 - 2^2)/ (1+ 4)

=> (-2 -2 -4 )/5

=> -8/5

**Therefore the required slope is -8/5**

The slope of the curve ,x^2+xy+y^2 = 7 at (1,2) is got by diffrentiating both sides of the equation and evaluating dy/dx or y' at (1,2).

x^2+xy+y^2 = 7

(x^2+xy+y^2)' = (7)'

2x+y+y' +2y*y' = 0

y'(1+2y) = -(2x+y)

y' = -(2x+y)/(1+2y)at(1,2) = -(2*1+2)/(1+2*2)

y' = -4/5

Slope = y' = -4/5 at (1,2).