Use implicit differentiation to find the slope of the tangent line to the curve ln(x+y)=x^2-y-89 at the point (9,-8).

We'll differentiate the expression of the given curve,
with respect to x, both sides:

d/dx[ln(x+y)] = d(x^2)/dx -
dy/dx - d(89)/dx

(1/(x+y))*d(x+y)/dx = 2x - y' -
0

(1 + y')/(x+y) = 2x - y'

1 +
y' = (x+y)(2x-y')

We'll remove the
brackets:

1 + y' = 2x^2 - x*y' + 2xy -
y*y'

We'll shift all the terms that contain y' to the left
side:

y' + x*y' + y*y' = 2x^2 + 2xy -
1

We'll factorize by y':

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

y' = (2x^2 + 2xy - 1)/(1 + x +
y)

We'll calculate the slope at the point
(9,-8):

y' = (162 - 144 -
1)/(1+9-8)

y' = 17/2

But y' =
m

The slope of the tangent line to the given
curve is m = 17/2.