Find y' x^2 + (y-x)^3=9 find an equation of the tangent line to the curve at (1,3)

Answer 1

Problem:

Find y' x^2 + (y-x)^3 = 9 find an equation of the tangent line to the curve at (1,3)

Part 1: Solve for y'

y' = dy/dx

Given: x^2 + (y - x)^3 = 9

Differentiating gives:

2x dx/dx + 3*(y - x)^2 * (y - x)' = d9/dx

2x dx/dx + 3*(y - x)^2 * (y' - dx/dx) = 0

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

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

Let's solve for y':

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

3*(y - x)^2 * (y' - 1) = -2x

(y' - 1) = -2x / [3*(y - x)^2]

y' = -2x / [3*(y - x)^2] + 1

y' = 1 - 2x / [3*(y - x)^2]

Part 2: Tangent Line:

point (1,3) = (x_0, y_0)

x = 1

y = 3

y' = 1 - 2(1) / [3*(3 - 1)^2]

y' = 1 - 2 / [3*(2)^2]

y' = 1 - 2 / [3*4]

m = 1 - 1 / 6

= 6/6 - 1 / 6

= 5/6

Equation of a Line:

y - y_0 = m(x - x_0))

y - 3 = 5/6(x - 1)

y = 5/6(x - 1) + 3

y = 5/6x -5/6 + 18/6

y = 5/6x + 13/6

y = 5/6x + 13/6

Answer 2

Differentiate implicitly given equation: 2x + 3(y - x)²(y' - 1) = 0, y' = -2x/[3(y- x)²] + 1;

Slope m of tangent line m = y' at (1,3); m = -2/12 + 1 = 5/6.

Equation of tangent line: y = 3 + 5/6(x - 1); y = 5/6x + 13/6