(arctan(sqrt(x)))^'

解答

(arctan (x))^{^{'}}

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

求解步骤

(arctan (x))^{'}

使用连锁律:

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

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

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

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

化简

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

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

\frac{d}{d x} (8 e^{x} + \frac{8}{3 x})

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

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x \to \infty lim (\frac{8 ^{x}}{9 ^{x + 1}})

\int \frac{32}{1 - cos ( 4 x )} d x