Question

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

Answer 1

I found `(dy)/(dx)=(e^xy-e^y)/(1+xe^y-e^x)`

Explanation:

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)`

Answer 2

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

Explanation:

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`