What is the (e^{-1/x})^' ?

The (e^{-1/x})^' is (e^{-1/x})/(x^2)

What is the first (e^{-1/x})^' ?

The first (e^{-1/x})^' is (e^{-1/x})/(x^2)

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(e^{- \frac{1}{x}})^{^{'}}

\frac{e ^{- \frac{1}{x}}}{x ^{2}}

Solution steps

(e^{- \frac{1}{x}})^{'}

Apply the chain rule:

e^{- \frac{1}{x}} (- \frac{1}{x})^{'}

= e^{- \frac{1}{x}} (- \frac{1}{x})^{'}

(- \frac{1}{x})^{'} = \frac{1}{x ^{2}}

= e^{- \frac{1}{x}} \frac{1}{x ^{2}}

Simplify

e^{- \frac{1}{x}} \frac{1}{x ^{2}} : \frac{e ^{- \frac{1}{x}}}{x ^{2}}

= \frac{e ^{- \frac{1}{x}}}{x ^{2}}