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}}

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

\frac{a ^{m}}{b ^{m}}

\frac{( - 99 ) ^{5}}{9 ^{5}}

5 e^{- (2)^{2}}