(1/((4-x)^2))^'

解答

(\frac{1}{( 4 - x ) ^{2}})^{^{'}}

\frac{2}{( 4 - x ) ^{3}}

求解步骤

(\frac{1}{( 4 - x ) ^{2}})^{'}

使用指数法则:

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

= ((4 - x)^{- 2})^{^{'}}

使用连锁律:

- \frac{2}{( 4 - x ) ^{3}} (4 - x)^{'}

= - \frac{2}{( 4 - x ) ^{3}} (4 - x)^{'}

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

= - \frac{2}{( 4 - x ) ^{3}} (- 1)

化简

= \frac{2}{( 4 - x ) ^{3}}

\int (\frac{sec ( x )}{1 + tan ( x )})^{2} d x

\frac{d}{d x} (e^{x s i n (3 x)})

\int_{0}^{5} \frac{2 t}{( t - 6 ) ^{2}} d t

\int \frac{x + 2}{x ^{2} + 6 x + 10} d x

x y^{^{'}} + 3 y = \frac{( e ^{x} )}{x}