We are using implicit differentiation to find the equation of the tangent to the

Question

We are using implicit differentiation to find the equation of the tangent to the curve
x3 + y = xy2 + 7 at the point (x,y)=(-2,-3).


We now know that y' satisfies the equation
y' = (3x*x - y*y)/(2xy - 1)

Substitute the coordinates of our point (x,y)=(-2,-3) into the equation and solve for y', the slope of the curve at the point (-2,-3).

y '(-2,-3) =

Explanation / Answer

Just double checking the implicit differentiation: 3x^2 + (dy/dx) = y^2 + 2xy(dy/dx) => 3x^2 - y^2 = (dy/dx)(2xy - 1) => y' = (3x^2 - y^2)/(2xy - 1) Okay, so it's correct :) Plug in the point (-2,-3) into y' to find the slope: Slope = (3(-2)^2 - (-3)^2)/(2(-2)(-3) - 1) = (3(4) - 9)/(12 - 1) = (12 - 9)/11 = 3/11 So we know the slope and the point it passes through. Using the point slope form, we can find the equation of the line: y - (-3) = (3/11)(x - (-2)) y + 3 = (3/11)(x + 2) y + 3 = (3/11)x + 6/11 y = (3/11)x + 6/11 - 3 = (3/11)x + 6/11 - 33/11 = (3/11)x - 27/11 So the equation of the tangent line is y = (3/11)x - 27/11

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