How do you implicitly differentiate #y = e^x y -x e^y#?

Question

How do you implicitly differentiate #y = e^x  y -x e^y#?

William Abdalla · Accepted Answer

I found #(dy)/(dx)=(e^xy-e^y)/(1+xe^y-e^x)# We consider #y# as a function of #x# and derive it as well to get: #1(dy)/(dx)=e^xy+e^x(dy)/(dx)-e^y-xe^y(dy)/(dx)# #(dy)/(dx)[e^x-xe^y-1]=e^y-e^xy# #(dy)/(dx)=(e^xy-e^y)/(1+xe^y-e^x)#

William Abdalla · Answer

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

differentiate using chain rule and product rule.

Differentiating with respect to x gives :

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

# dy/dx = e^x dy/dx+ye^x - ( x.e^y .d/dx(y) + e^y.1 )#

# dy/dx = e^xdy/dx+ye^x - ( x.e^y.dy/dx + e^y )#

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

collecting# dy/dx# terms to left hand side gives :

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

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

# rArr dy/dx = (ye^x - e^y )/(1 - e^x + xe^y #

Paisley Johnson · Answer

undefined To implicitly differentiate $ y = e^x \cdot y - x \cdot e^y $, we follow these steps:

1. Differentiate both sides of the equation with respect to $ x $.
2. Use the product rule when differentiating terms that contain both $ x $ and $ y $.
3. Remember that $ \frac{dy}{dx} $ represents $ \frac{dy}{dx} $.

The implicit differentiation of the given equation yields:

$$ \frac{dy}{dx} = e^x \cdot \frac{dy}{dx} + e^x \cdot y - \left(1 + x \cdot \frac{dy}{dx} \cdot e^y\right) $$

Now, we isolate $ \frac{dy}{dx} $:

$$ \frac{dy}{dx} - e^x \cdot \frac{dy}{dx} = e^x \cdot y - \left(1 + x \cdot \frac{dy}{dx} \cdot e^y\right) $$

$$ \frac{dy}{dx} \left(1 - e^x\right) + x \cdot \frac{dy}{dx} \cdot e^y = e^x \cdot y - 1 $$

$$ \frac{dy}{dx} \left(1 - e^x + x \cdot e^y\right) = e^x \cdot y - 1 $$

$$ \frac{dy}{dx} = \frac{e^x \cdot y - 1}{1 - e^x + x \cdot e^y} $$