Derivative of $$$e^{- \frac{1}{u^{2}}}$$$
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Find $$$\frac{d}{du} \left(e^{- \frac{1}{u^{2}}}\right)$$$.
Solution
The function $$$e^{- \frac{1}{u^{2}}}$$$ is the composition $$$f{\left(g{\left(u \right)} \right)}$$$ of two functions $$$f{\left(v \right)} = e^{v}$$$ and $$$g{\left(u \right)} = - \frac{1}{u^{2}}$$$.
Apply the chain rule $$$\frac{d}{du} \left(f{\left(g{\left(u \right)} \right)}\right) = \frac{d}{dv} \left(f{\left(v \right)}\right) \frac{d}{du} \left(g{\left(u \right)}\right)$$$:
$${\color{red}\left(\frac{d}{du} \left(e^{- \frac{1}{u^{2}}}\right)\right)} = {\color{red}\left(\frac{d}{dv} \left(e^{v}\right) \frac{d}{du} \left(- \frac{1}{u^{2}}\right)\right)}$$The derivative of the exponential is $$$\frac{d}{dv} \left(e^{v}\right) = e^{v}$$$:
$${\color{red}\left(\frac{d}{dv} \left(e^{v}\right)\right)} \frac{d}{du} \left(- \frac{1}{u^{2}}\right) = {\color{red}\left(e^{v}\right)} \frac{d}{du} \left(- \frac{1}{u^{2}}\right)$$Return to the old variable:
$$e^{{\color{red}\left(v\right)}} \frac{d}{du} \left(- \frac{1}{u^{2}}\right) = e^{{\color{red}\left(- \frac{1}{u^{2}}\right)}} \frac{d}{du} \left(- \frac{1}{u^{2}}\right)$$Apply the constant multiple rule $$$\frac{d}{du} \left(c f{\left(u \right)}\right) = c \frac{d}{du} \left(f{\left(u \right)}\right)$$$ with $$$c = -1$$$ and $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$:
$$e^{- \frac{1}{u^{2}}} {\color{red}\left(\frac{d}{du} \left(- \frac{1}{u^{2}}\right)\right)} = e^{- \frac{1}{u^{2}}} {\color{red}\left(- \frac{d}{du} \left(\frac{1}{u^{2}}\right)\right)}$$Apply the power rule $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ with $$$n = -2$$$:
$$- e^{- \frac{1}{u^{2}}} {\color{red}\left(\frac{d}{du} \left(\frac{1}{u^{2}}\right)\right)} = - e^{- \frac{1}{u^{2}}} {\color{red}\left(- \frac{2}{u^{3}}\right)}$$Thus, $$$\frac{d}{du} \left(e^{- \frac{1}{u^{2}}}\right) = \frac{2 e^{- \frac{1}{u^{2}}}}{u^{3}}$$$.
Answer
$$$\frac{d}{du} \left(e^{- \frac{1}{u^{2}}}\right) = \frac{2 e^{- \frac{1}{u^{2}}}}{u^{3}}$$$A