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

2 Answers
Nov 13, 2016

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#

Nov 13, 2016

#-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#

enter image source here

and the second is associated to #xy = 1#

enter image source here

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.