# 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).

justaguide | College Teacher | (Level 2) Distinguished Educator

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The slope of the tangent to the graph ln(x+y)=x^2-y-89 at the point (9,-8) is the value of the first derivative dy/dx at that point.

[ln(x+y)]' = [x^2 - y - 89]'

=> (1 + dy/dx)/(x + y) = 2x - dy/dx

=> 1 + dy/dx = (2x - dy/dx)(x + y)

=> 1 + dy/dx = 2x(x + y) - dy/dx(x + y)

=> dy/dx[ 1 + x + y] = 2x(x + y) - 1

=> dy/dx = [2x(x + y) - 1]/[ 1 + x + y]

At the point (9, -8)

dy/dx = [2*-8*1 - 1]/2

=> -17/2

=> -8.5

The slope if the required tangent is -8.5

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.