derivative of 1/(1+e^{1/x})

解答

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

\frac{e ^{\frac{1}{x}}}{x ^{2} ( 1 + e ^{\frac{1}{x}} ) ^{2}}

求解步骤

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

使用指数法则:

\frac{1}{a} = a^{- 1}

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

使用连锁律:

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

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

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

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

化简

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

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

x \to 3 - lim (\frac{∣ x ∣}{x})

- x y + (3 + x) (\frac{d y}{d x}) = 0

n = 1 \sum \infty 8 (0.8)^{n - 1}

\int \frac{5 x ^{2} - 3 x + 2}{x ^{3} - 2 x ^{2}} d x

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