find the equation of the tangent line, at the designated point x³+y³=6xy, P(3,3)

Answer 1

x³ + y³ = 6xy

3x² + 3y²(dy/dx) = 6[y + x(dy/dx)]

At (3,3), we have 27 + 27(dy/dx) = 6[3 + 3 (dy/dx)]

27 + 27(dy/dx) = 18 + 18(dy/dx)

9(dy/dx) = -9

So, dy/dx = -1

Equation of tangent line: y - 3 = -1(x - 3)

y = -x + 6

Answer 2

Since you can’t isolate y to differentiate, you need to use implicit differentiation. So any terms with y, we must consider the derivative of y, so it needs to have a dy/dx in its derivative. Additionally, the right hand side we also need to implement the product rule.

3x² + 3y² (dy/dx) = 6x(1)(dy/dx) + y(6)

Now we will move all terms with a dy/dx to one side.

3y²(dy/dx) - 6x(dy/dx) = 6y - 3x²

Next, we will factor a dy/dx out of the terms on the left hand side and then isolate dy/dx.

(dy/dx)[3y² - 6x] = 6y - 3x²

dy/dx = [6y - 3x²] / [3y² - 6x]

Now we have the derivative, dy/dx, which is the slope of the tangent line at any point. Let’s now substitute in the point (3,3) to get the slope of the tangent.

dy/dx = [6(3)-3(3)²] / [3(3)² -6(3)]

dy/dx = -9 / 9

dy/dx = -1

To get the tangent line in the form y = mx + b we can now substitute our slope -1 for m and our coordinate pair (3,3) for x and y in order to solve for b

3 = -1(3) + b

b = 6

So the equation of the tangent line is

y = -x + 6