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}}

f (x) = (x^{2} + 2 x)^{3} (7 x + 1)^{4}

\frac{d y}{d x} = - \frac{x}{y}

\frac{d}{d x} (2 sin (x) + 4 cos (x) sin (x))

\int \frac{1}{( x ^{2} ) \cdot 25 x ^{2} + 16} d x

\frac{\partial}{\partial y} (x y + 1)