Find the Tangent Line at the Point x^2+xy+y^2=3 , (1,1)

Problem

x2+x*y+y2=3,(1,1)

Solution

1. Differentiate implicitly with respect to x by applying the derivative to both sides of the equation.

d(x2)/d(x)+(d(x)*y)/d(x)+d(y2)/d(x)=d(3)/d(x)

2. Apply the product rule to the term x*y and the chain rule to the term y2

2*x+(xd(y)/d(x)+y)+2*yd(y)/d(x)=0

3. Isolate the derivative d(y)/d(x) by moving terms without d(y)/d(x) to the right side and factoring.

xd(y)/d(x)+2*yd(y)/d(x)=−2*x−y

d(y)/d(x)*(x+2*y)=−2*x−y

d(y)/d(x)=(−2*x−y)/(x+2*y)

4. Calculate the slope m by substituting the given point (1,1) into the expression for d(y)/d(x)

m=(−2*(1)−1)/(1+2*(1))

m=(−3)/3=−1

5. Use the point-slope formula y−(y_1)=m*(x−(x_1)) with m=−1 and the point (1,1) to find the equation of the tangent line.

y−1=−1*(x−1)

y−1=−x+1

y=−x+2

Final Answer

y=−x+2

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