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

Answer 1

#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!

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Answer 2

#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.!

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Answer 3

To implicitly differentiate the equation ( -y = xy - xe^{-y} ), follow these steps:

  1. Differentiate both sides of the equation with respect to ( x ).
  2. Apply the product rule when differentiating terms involving ( y ).
  3. Use the chain rule when differentiating ( e^{-y} ).

Differentiating ( -y ) with respect to ( x ) gives ( -\frac{dy}{dx} ).

For the term ( xy ), apply the product rule: ( \frac{d(xy)}{dx} = x\frac{dy}{dx} + y ).

For the term ( xe^{-y} ), differentiate ( x ) with respect to ( x ) (which gives 1), then differentiate ( e^{-y} ) using the chain rule: ( \frac{d(e^{-y})}{dx} = -e^{-y} \frac{dy}{dx} ). So, ( \frac{d(xe^{-y})}{dx} = 1 \cdot e^{-y} + x \cdot (-e^{-y}) \frac{dy}{dx} ).

Combine these results to get:

[ -\frac{dy}{dx} = x\frac{dy}{dx} + y - e^{-y} + x(-e^{-y})\frac{dy}{dx} ]

Rearrange the terms and solve for ( \frac{dy}{dx} ):

[ -\frac{dy}{dx} - x\frac{dy}{dx} - x(-e^{-y})\frac{dy}{dx} = y - e^{-y} ]

[ (-1 - x - xe^{-y})\frac{dy}{dx} = y - e^{-y} ]

[ \frac{dy}{dx} = \frac{y - e^{-y}}{-1 - x - xe^{-y}} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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