Answer:
The first derivative of [tex]y = x\cdot e^{y}[/tex] is [tex]y' = \frac{e^{y}}{1-x\cdot e^{y}}[/tex].
Step-by-step explanation:
Let [tex]y = x\cdot e^{y}[/tex], we apply natural logarithms in both sides of the expression:
[tex]\ln y = \ln x\cdot e^{y}[/tex]
[tex]\ln y = \ln x +\ln e^{y}[/tex]
[tex]\ln y = \ln x +y[/tex]
[tex]\ln y -y = \ln x[/tex] (1)
Then, we differentiate (1) respect to [tex]x[/tex]:
[tex]\frac{y'}{y}-y' = \frac{1}{x}[/tex]
[tex]y'\cdot \left(\frac{1}{y}-1\right) = \frac{1}{x}[/tex]
[tex]y'\cdot \left(\frac{1-y}{y} \right) = \frac{1}{x}[/tex]
[tex]y' = \frac{1}{x}\cdot \left(\frac{y}{1-y} \right)[/tex]
[tex]y' = \frac{1}{x}\cdot \left(\frac{x\cdot e^{y}}{1-x\cdot e^{y}} \right)[/tex]
[tex]y' = \frac{e^{y}}{1-x\cdot e^{y}}[/tex]
