How do you implicitly differentiate #-y=xy-xe^-y #?

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Answer 1 · 2017-02-02T07:18:12

#dy/dx=y/{x(x+1)(y+1)}.#

Since, the Problem is to be solved using Implicit Diffn., we will

first solve it accordingly.

#-y=xy-xe^-y rArr xe^-y=xy+y#

#:. d/dx(xe^-y)=d/dx(xy+y)=d/dx(xy)+d/dx(y)......................................"[Addition Rule]"#

By the Product Rule , then,

#xd/dxe^-y+e^-yd/dx(x)={xd/dx(y)+yd/dx(x)}+dy/dx,# i.e.,

#xd/dxe^-y+e^-y=xdy/dx+y+dy/dx... (star)#

To find #d/dx(e^-y),# we use the Chain Rule to get,

#d/dx(e^-y)=d/dy(e^-y)d/dx(-y)=(e^-y)(-dy/dx).#

Hence, continuing from #(star)#,

#-xe^-ydy/dx+e^-y=xdy/dx+y+dy/dx#

#:. e^-y-y=(xe^-y+x+1)dy/dx#

#rArr dy/dx=(e^-y-y)/(xe^-y+x+1)," &, as "e^-y=(xy+y)/x,# we have,

# dy/dx={(xy+y)/x-y}/(xy+y+x+1)=y/[x{y(x+1)+1(x+1)}], i.e., #

#dy/dx=y/{x(x+1)(y+1)}.#

Enjoy Maths.!

Answer 2 · 2017-02-02T07:40:24

#dy/dx=y/{x(x+1)(y+1)}.#

As a Second Method , we can solve the Problem without the

use of Implicit Diffn.

Let us rewrite the given eqn. as, #y=xe^-y-xy=x(e^-y-y),#

giving, #x=y/(e^-y-y).#

Diff.ing w.r.t. #y# using the Quotient Rule and the Chain Rule,

#dx/dy={(e^-y-y)d/dy(y)-yd/dy(e^-y-y)}/(e^-y-y)^2,# i.e.,

#dx/dy=[(e^-y-y)-y{(e^-y)(-1)-1}]/(e^-y-y)^2#

#=(e^-y-y+ye^-y+y)/(e^-y-y)^2#

#:. dx/dy=((y+1)e^-y)/(e^-y-y)^2#

#"Therefore, "dy/dx=(e^-y-y)^2/((y+1)e^-y).#

I leave it, as an exercise, to show that our both Answers match!