Use implicit differentiation to find an equation of the tangent line to the curve at the given point. y2 (6 − x) = x3, 2,sqrt2 , (cissoid of Diocles)

Answer 1

y²(6 - x) = x³

The left side of the equation y²(6 - x) is the product of two things so we must use the product rule:

(u•v)' = u'v + uv' where u = y² then u' = 2y•dy/dx and v = (6 - x) so v' = -1. Putting these together:

[y²(6 - x)]' = (2y•dy/dx)(6 - x) + (y²)(-1) = dy/dx(12y - 2xy) - y²

The right side of the equation is a simple use of the power rule (x³)' = 3x²

So:

dy/dx(12y - 2xy) - y² = 3x²

dy/dx(12y - 2xy) = 3x² + y²

dy/dx = (3x² + y²)/(12y - 2xy)

So the slope of the tangent line is found by plugging in (2, √2) or x = 2, y = √2

dy/dx at (2, √2) = (3(2)² + (√2)²)/(12(√2) - 2(2)(√2)) = (12 + 2)/(8√2) = 14/(8√2) = 7/(4√2) = 7√2/8

So m = 7√2/8

Using the point slope form, y - y₁ = m(x - x₁), we get:

y - √2 = 7√2/8(x - 2)