What is the d/(ds)(ln((s-1)/(s^2+2s+5))) ?

The d/(ds)(ln((s-1)/(s^2+2s+5))) is (-s^2+2s+7)/((s-1)(s^2+2s+5))

What is the first d/(ds)(ln((s-1)/(s^2+2s+5))) ?

The first d/(ds)(ln((s-1)/(s^2+2s+5))) is (-s^2+2s+7)/((s-1)(s^2+2s+5))

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\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 )}

Solution steps

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

Apply the chain rule:

\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}}

Simplify

\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 )}

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

What is the d/(ds)(ln((s-1)/(s^2+2s+5))) ?
The d/(ds)(ln((s-1)/(s^2+2s+5))) is (-s^2+2s+7)/((s-1)(s^2+2s+5))
What is the first d/(ds)(ln((s-1)/(s^2+2s+5))) ?
The first d/(ds)(ln((s-1)/(s^2+2s+5))) is (-s^2+2s+7)/((s-1)(s^2+2s+5))