d/(ds)(ln((s-1)/(s^2+2s+5)))

解答

\frac{d}{d s} (ln (\frac{s - 1}{s ^{2} + 2 s + 5}))

\frac{- s ^{2} + 2 s + 7}{( s - 1 ) ( s ^{2} + 2 s + 5 )}

求解步骤

\frac{d}{d s} (ln (\frac{s - 1}{s ^{2} + 2 s + 5}))

使用连锁律:

\frac{1}{\frac{s - 1}{s ^{2} + 2 s + 5}} \frac{d}{d s} (\frac{s - 1}{s ^{2} + 2 s + 5})

= \frac{1}{\frac{s - 1}{s ^{2} + 2 s + 5}} \frac{d}{d s} (\frac{s - 1}{s ^{2} + 2 s + 5})

\frac{d}{d s} (\frac{s - 1}{s ^{2} + 2 s + 5}) = \frac{- s ^{2} + 2 s + 7}{( s ^{2} + 2 s + 5 ) ^{2}}

= \frac{1}{\frac{s - 1}{s ^{2} + 2 s + 5}} \cdot \frac{- s ^{2} + 2 s + 7}{( s ^{2} + 2 s + 5 ) ^{2}}

化简

\frac{1}{\frac{s - 1}{s ^{2} + 2 s + 5}} \cdot \frac{- s ^{2} + 2 s + 7}{( s ^{2} + 2 s + 5 ) ^{2}} : \frac{- s ^{2} + 2 s + 7}{( s - 1 ) ( s ^{2} + 2 s + 5 )}

= \frac{- s ^{2} + 2 s + 7}{( s - 1 ) ( s ^{2} + 2 s + 5 )}

t an g e n t f (x) = x^{2} - x + 2, at x = - 1

9 y^{^{''}} + 18 y^{^{'}} + 17 y = 0

n = 1 \sum \infty \frac{2 n}{5 n + 1}

\frac{\partial}{\partial y} (\frac{12 x}{x + y})

\frac{d y}{d x} = y - 2