Question

Find an equation for the tangent line to the curve x^3+xy+y^3=x at the point (1,0)
Would the first step be x^3+y+x+3y^2=x

Answer 1

No, you need to take the derivative, you have to take the derivative of each item, not just the y's:

\[d/dx(x^3+xy+y^3)=d/dx(x)\]
\[3x^2 + 1y + xy' + 3y^2y'=1\]
\[xy'+3y^2y'=1-3x^2-y\]
\[y'(x+3y^2)=1-3x^2-y\]
\[y'=(1-3x^2-y)/(x+3y^2)\]

Now, put (1, 0) into y' to get the slope of the tangent line. Put a 1 where you see an x and a 0 where you see a y.

y' = (1 - 3(1)^2-0)/(1+3(0)^2) = -2/1 = -2

Now that you have a slope m = -2, use that with the point (1, 0) and get an equation in point-slope form: y - 0 = -2(x - 1) --> y = -2x + 2

Answer 2

Thank You