derivative (e^{1/x})/(x^3)

Lösung

ableitung von

\frac{e ^{\frac{1}{x}}}{x ^{3}}

- \frac{e ^{\frac{1}{x}} ( 3 x + 1 )}{x ^{5}}

Schritte zur Lösung

\frac{d}{d x} (\frac{e ^{\frac{1}{x}}}{x ^{3}})

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( e ^{\frac{1}{x}} ) x ^{3} - \frac{d}{d x} ( x ^{3} ) e ^{\frac{1}{x}}}{( x ^{3} ) ^{2}}

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

\frac{d}{d x} (x^{3}) = 3 x^{2}

= \frac{( - \frac{e ^{\frac{1}{x}}}{x ^{2}} ) x ^{3} - 3 x ^{2} e ^{\frac{1}{x}}}{( x ^{3} ) ^{2}}

Vereinfache

\frac{( - \frac{e ^{\frac{1}{x}}}{x ^{2}} ) x ^{3} - 3 x ^{2} e ^{\frac{1}{x}}}{( x ^{3} ) ^{2}} : - \frac{e ^{\frac{1}{x}} ( 3 x + 1 )}{x ^{5}}

= - \frac{e ^{\frac{1}{x}} ( 3 x + 1 )}{x ^{5}}

t an g e n t f (x) = x + 1, (3, 2)

\int C e^{- a t} d t

d er i v a t i v e f (x) = \frac{6}{x ^{4}}

y^{^{'}} = \frac{x + 2 y}{x}

\int e^{- l n (x)} d x