Question

Y=xe^y differentiate by logarithmic method​

Answer 1

The first derivative of
`y = x\cdot e^(y)` is
`y' = (e^(y))/(1-x\cdot e^(y))`.

Explanation:

Let
`y = x\cdot e^(y)`, we apply natural logarithms in both sides of the expression:

`\ln y = \ln x\cdot e^(y)`

`\ln y = \ln x +\ln e^(y)`

`\ln y = \ln x +y`

`\ln y -y = \ln x` (1)

Then, we differentiate (1) respect to
`x`:

`(y')/(y)-y' = (1)/(x)`

`y'\cdot \left((1)/(y)-1\right) = (1)/(x)`

`y'\cdot \left((1-y)/(y) \right) = (1)/(x)`

`y' = (1)/(x)\cdot \left((y)/(1-y) \right)`

`y' = (1)/(x)\cdot \left((x\cdot e^(y))/(1-x\cdot e^(y)) \right)`

`y' = (e^(y))/(1-x\cdot e^(y))`