Question

Y=xe^y differentiate by logarithmic method​

Answer 1

Answer:

The first derivative of $y = x\cdot e^{y}$ is $y' = \frac{e^{y}}{1-x\cdot e^{y}}$.

Step-by-step 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 $x$:

$\frac{y'}{y}-y' = \frac{1}{x}$

$y'\cdot \left(\frac{1}{y}-1\right) = \frac{1}{x}$

$y'\cdot \left(\frac{1-y}{y} \right) = \frac{1}{x}$

$y' = \frac{1}{x}\cdot \left(\frac{y}{1-y} \right)$

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

$y' = \frac{e^{y}}{1-x\cdot e^{y}}$