Implicit derivative of $$$x e^{y} = x - y$$$ with respect to $$$x$$$
Your Input
Find $$$\frac{d}{dx} \left(x e^{y} = x - y\right)$$$.
Solution
Differentiate separately both sides of the equation (treat $$$y$$$ as a function of $$$x$$$): $$$\frac{d}{dx} \left(x e^{y{\left(x \right)}}\right) = \frac{d}{dx} \left(x - y{\left(x \right)}\right)$$$.
Differentiate the LHS of the equation.
Apply the product rule $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ with $$$f{\left(x \right)} = x$$$ and $$$g{\left(x \right)} = e^{y{\left(x \right)}}$$$:
$${\color{red}\left(\frac{d}{dx} \left(x e^{y{\left(x \right)}}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x\right) e^{y{\left(x \right)}} + x \frac{d}{dx} \left(e^{y{\left(x \right)}}\right)\right)}$$The function $$$e^{y{\left(x \right)}}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = e^{u}$$$ and $$$g{\left(x \right)} = y{\left(x \right)}$$$.
Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$$x {\color{red}\left(\frac{d}{dx} \left(e^{y{\left(x \right)}}\right)\right)} + e^{y{\left(x \right)}} \frac{d}{dx} \left(x\right) = x {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(y{\left(x \right)}\right)\right)} + e^{y{\left(x \right)}} \frac{d}{dx} \left(x\right)$$The derivative of the exponential is $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:
$$x {\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + e^{y{\left(x \right)}} \frac{d}{dx} \left(x\right) = x {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + e^{y{\left(x \right)}} \frac{d}{dx} \left(x\right)$$Return to the old variable:
$$x e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + e^{y{\left(x \right)}} \frac{d}{dx} \left(x\right) = x e^{{\color{red}\left(y{\left(x \right)}\right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + e^{y{\left(x \right)}} \frac{d}{dx} \left(x\right)$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$x e^{y{\left(x \right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + e^{y{\left(x \right)}} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = x e^{y{\left(x \right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + e^{y{\left(x \right)}} {\color{red}\left(1\right)}$$Simplify:
$$x e^{y{\left(x \right)}} \frac{d}{dx} \left(y{\left(x \right)}\right) + e^{y{\left(x \right)}} = \left(x \frac{d}{dx} \left(y{\left(x \right)}\right) + 1\right) e^{y{\left(x \right)}}$$Thus, $$$\frac{d}{dx} \left(x e^{y{\left(x \right)}}\right) = \left(x \frac{d}{dx} \left(y{\left(x \right)}\right) + 1\right) e^{y{\left(x \right)}}$$$.
Differentiate the RHS of the equation.
The derivative of a sum/difference is the sum/difference of derivatives:
$${\color{red}\left(\frac{d}{dx} \left(x - y{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x\right) - \frac{d}{dx} \left(y{\left(x \right)}\right)\right)}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$${\color{red}\left(\frac{d}{dx} \left(x\right)\right)} - \frac{d}{dx} \left(y{\left(x \right)}\right) = {\color{red}\left(1\right)} - \frac{d}{dx} \left(y{\left(x \right)}\right)$$Thus, $$$\frac{d}{dx} \left(x - y{\left(x \right)}\right) = 1 - \frac{d}{dx} \left(y{\left(x \right)}\right)$$$.
Therefore, we have obtained the following linear equation with respect to the derivative: $$$\left(x \frac{dy}{dx} + 1\right) e^{y} = 1 - \frac{dy}{dx}$$$.
Solving it, we obtain that $$$\frac{dy}{dx} = \frac{1 - e^{y}}{x e^{y} + 1}$$$.
Answer
$$$\frac{dy}{dx} = \frac{1 - e^{y}}{x e^{y} + 1}$$$A