If ln(x^(2)-3y)=x-y+3

Question

y(-2)=1, find y prime(-2). Also find the slope of the tangent line at (-2,1).

Answer 1

We are told that (x,y) = (-2, 1) is a solution. Let's verify this.

ln(x² - 3y) = x - y + 3

ln[(-2)² - 3(1)] = -2 -1 + 3

ln(4 - 3) = 0

ln(1) = 0, which is true. So, check.

Let us perform implicit differentiation.

[2x - 3 dy/dx]/[x² - 3y] = 1 - dy/dx

Cross multiply

2x - 3 dy/dx = (x² - 3y)(1 - dy/dx)

We want the tangent to the curve at the point (x,y) = (-2, 1). So, let x = -2 and y = 1.

2(-2) - 3 dy/dx = [(-2)² - 3(1)](1 - dy/dx)

-4 - 3 dy/dx = [4 - 3](1 - dy/dx)

-4 - 3 dy/dx = 1 - dy/dx

- 2 dy/dx = 5

dy/dx = -5/2 at (x, y) = (-2, 1)