Derivative of $$$e^{\frac{1}{x}}$$$
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Solution
The function $$$e^{\frac{1}{x}}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = e^{u}$$$ and $$$g{\left(x \right)} = \frac{1}{x}$$$.
Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(e^{\frac{1}{x}}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(\frac{1}{x}\right)\right)}$$The derivative of the exponential is $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:
$${\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(\frac{1}{x}\right) = {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(\frac{1}{x}\right)$$Return to the old variable:
$$e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(\frac{1}{x}\right) = e^{{\color{red}\left(\frac{1}{x}\right)}} \frac{d}{dx} \left(\frac{1}{x}\right)$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = -1$$$:
$$e^{\frac{1}{x}} {\color{red}\left(\frac{d}{dx} \left(\frac{1}{x}\right)\right)} = e^{\frac{1}{x}} {\color{red}\left(- \frac{1}{x^{2}}\right)}$$Thus, $$$\frac{d}{dx} \left(e^{\frac{1}{x}}\right) = - \frac{e^{\frac{1}{x}}}{x^{2}}$$$.
Answer
$$$\frac{d}{dx} \left(e^{\frac{1}{x}}\right) = - \frac{e^{\frac{1}{x}}}{x^{2}}$$$A