Answer: The required value of [tex]\dfrac{dy}{dx}[/tex] is [tex]\dfrac{y}{1-y}.[/tex]
Step-by-step explanation: We are given to find the value of [tex]\dfrac{dy}{dx}[/tex] from the following equation :
[tex]e^{x+y}=y~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
Taking natural logarithm on both sides of equation (i), we have
[tex]\ln e^{x+y}=\ln y\\\\\Rightarrow x+y=\ln y~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]
Differentiating both sides of equation (ii) with respect to x, we get
[tex]\dfrac{d}{dx}(x+y)=\dfrac{d}{dx}\ln y\\\\\\\Rightarrow 1+\dfrac{dy}{dx}=\dfrac{1}{y}\dfrac{dy}{dx}\\\\\\\Rightarrow \left(\dfrac{1}{y}-1\right)\dfrac{dy}{dx}=1\\\\\\\Rightarrow \dfrac{1-y}{y}\dfrac{dy}{dx}=1\\\\\\\Rightarrow \dfrac{dy}{dx}=\dfrac{y}{1-y}.[/tex]
Thus, the required value of [tex]\dfrac{dy}{dx}[/tex] is [tex]\dfrac{y}{1-y}.[/tex]
