(ln(x+2))^'

解答

(ln (x + 2))^{^{'}}

\frac{1}{x + 2}

求解步骤

(ln (x + 2))^{'}

使用连锁律:

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

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

(x + 2)^{'} = 1

= \frac{1}{x + 2} \cdot 1

化简

= \frac{1}{x + 2}

x \to 7 - lim (\frac{6 x}{x - 7})

\frac{d}{d x} (- 1.2 x^{2} + 9.6 x)

in v erse l a pl a ce \frac{( 1 - e ^{- 5 s} )}{s ^{2}}

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