(1/(1+x))^'

解答

(\frac{1}{1 + x})^{^{'}}

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

求解步骤

(\frac{1}{1 + x})^{'}

使用指数法则:

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

= ((1 + x)^{- 1})^{^{'}}

使用连锁律:

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

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

(1 + x)^{'} = 1

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

化简

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

x \to 0 lim (1 + csc (x))

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