d/(ds)(6/((s-2)^2+36))

解答

\frac{d}{d s} (\frac{6}{( s - 2 ) ^{2} + 36})

- \frac{6 ( 2 s - 4 )}{( s ^{2} - 4 s + 40 ) ^{2}}

求解步骤

\frac{d}{d s} (\frac{6}{( s - 2 ) ^{2} + 36})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 6 \frac{d}{d s} (\frac{1}{( s - 2 ) ^{2} + 36})

使用指数法则:

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

= 6 \frac{d}{d s} (((s - 2)^{2} + 36)^{- 1})

使用连锁律:

- \frac{1}{( ( s - 2 ) ^{2} + 36 ) ^{2}} \frac{d}{d s} ((s - 2)^{2} + 36)

= - \frac{1}{( ( s - 2 ) ^{2} + 36 ) ^{2}} \frac{d}{d s} ((s - 2)^{2} + 36)

\frac{d}{d s} ((s - 2)^{2} + 36) = 2 s - 4

= 6 - \frac{1}{( ( s - 2 ) ^{2} + 36 ) ^{2}} (2 s - 4)

化简

6 - \frac{1}{( ( s - 2 ) ^{2} + 36 ) ^{2}} (2 s - 4) : - \frac{6 ( 2 s - 4 )}{( s ^{2} - 4 s + 40 ) ^{2}}

= - \frac{6 ( 2 s - 4 )}{( s ^{2} - 4 s + 40 ) ^{2}}

\int \frac{1}{x ^{5} + x ^{4} + 2 x ^{3} + 2 x ^{2} + x + 1} d x

l a pl a ce t r an s f or m 2 cos (t - \frac{π}{6})

\int_{0}^{\infty} e^{- y} d y

\int 7 x + 8 y^{2} d x

\int \frac{1}{( x - 5 ) ^{2} ( x + 3 )} d x