Derivative of $$$x e^{- x}$$$
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Solution
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 $$$g{\left(x \right)} = e^{- x}$$$:
$${\color{red}\left(\frac{d}{dx} \left(x e^{- x}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x\right) e^{- x} + x \frac{d}{dx} \left(e^{- x}\right)\right)}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$x \frac{d}{dx} \left(e^{- x}\right) + e^{- x} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = x \frac{d}{dx} \left(e^{- x}\right) + e^{- x} {\color{red}\left(1\right)}$$The function $$$e^{- x}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = e^{u}$$$ and $$$g{\left(x \right)} = - x$$$.
Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$$x {\color{red}\left(\frac{d}{dx} \left(e^{- x}\right)\right)} + e^{- x} = x {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(- x\right)\right)} + e^{- x}$$The derivative of the exponential is $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:
$$x {\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(- x\right) + e^{- x} = x {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(- x\right) + e^{- x}$$Return to the old variable:
$$x e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(- x\right) + e^{- x} = x e^{{\color{red}\left(- x\right)}} \frac{d}{dx} \left(- x\right) + e^{- x}$$Apply the constant multiple rule $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ with $$$c = -1$$$ and $$$f{\left(x \right)} = x$$$:
$$x e^{- x} {\color{red}\left(\frac{d}{dx} \left(- x\right)\right)} + e^{- x} = x e^{- x} {\color{red}\left(- \frac{d}{dx} \left(x\right)\right)} + e^{- x}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$- x e^{- x} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + e^{- x} = - x e^{- x} {\color{red}\left(1\right)} + e^{- x}$$Simplify:
$$- x e^{- x} + e^{- x} = \left(1 - x\right) e^{- x}$$Thus, $$$\frac{d}{dx} \left(x e^{- x}\right) = \left(1 - x\right) e^{- x}$$$.
Answer
$$$\frac{d}{dx} \left(x e^{- x}\right) = \left(1 - x\right) e^{- x}$$$A