Question

How do you use implicit differentiation to find the slope of the curve given `1/(x+1)+1/(y+1)=1` at (1,1)?

Answer 1

The curve has a slope of `-1`.

Explanation:

`(0(x + 1) - 1 xx 1)/(x + 1)^2 + (0(y + 1) - 1 xx 1(dy/dx))/(y + 1)^2 = 0`

`-1/(x + 1)^2 - (1(dy/dx))/(y + 1)^2 = 0`

Solve for `dy/dx`.

`-((y + 1)^2 + (x + 1)^2dy/dx)/((x + 1)^2(y + 1)^2) = 0`

`-(y + 1)^2 - (x + 1)^2dy/dx = 0`

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

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

The slope, or rate of change of the curve is

`dy/dx=-(1 + 1)^2/(1 + 1)^2`

`dy/dx= -4/4`

`dy/dx = -1`

Answer 2

`-1`

Explanation:

`1/(x+1)+1/(y+1)=(x+y+2)/(xy+x+y+1)=1`

then

`x+y+2=xy+x+y+1` simplifying

`xy=1` now

`d/(dx)(xy-1)=y+xy'=0` so

`y'=-1/x^2` so `y'(1)=-1`

NOTE. What is de difference between `1/(x+1)+1/(y+1)=1` and `xy=1`?

Attached two graphics. The first is associated to `1/(x+1)+1/(y+1)=1`

and the second is associated to `xy = 1`

This second graphic is clean, without the singularities in `x=-1` and `y=-1`. If the desired results are asked apart from the singularities, this second option offers a terse environment to the problem.