What is the d/(ds)((s+2)/(s^2+4s+13)) ?

The d/(ds)((s+2)/(s^2+4s+13)) is (-s^2-4s+5)/((s^2+4s+13)^2)

What is the first d/(ds)((s+2)/(s^2+4s+13)) ?

The first d/(ds)((s+2)/(s^2+4s+13)) is (-s^2-4s+5)/((s^2+4s+13)^2)

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\frac{d}{d s} (\frac{s + 2}{s ^{2} + 4 s + 13})

\frac{- s ^{2} - 4 s + 5}{( s ^{2} + 4 s + 13 ) ^{2}}

Solution steps

\frac{d}{d s} (\frac{s + 2}{s ^{2} + 4 s + 13})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d s} ( s + 2 ) ( s ^{2} + 4 s + 13 ) - \frac{d}{d s} ( s ^{2} + 4 s + 13 ) ( s + 2 )}{( s ^{2} + 4 s + 13 ) ^{2}}

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

\frac{d}{d s} (s^{2} + 4 s + 13) = 2 s + 4

= \frac{1 \cdot ( s ^{2} + 4 s + 13 ) - ( 2 s + 4 ) ( s + 2 )}{( s ^{2} + 4 s + 13 ) ^{2}}

Simplify

\frac{1 \cdot ( s ^{2} + 4 s + 13 ) - ( 2 s + 4 ) ( s + 2 )}{( s ^{2} + 4 s + 13 ) ^{2}} : \frac{- s ^{2} - 4 s + 5}{( s ^{2} + 4 s + 13 ) ^{2}}

= \frac{- s ^{2} - 4 s + 5}{( s ^{2} + 4 s + 13 ) ^{2}}

f (x) = x^{2} + 6, (5, 31)

\int e^{- t^{2}} d t

\int \frac{1}{z ^{2} a ^{2} - x ^{2}} d z

x^{3} \frac{16}{3}

f (x) = 2 sin (x) + sin^{2} (x)

What is the d/(ds)((s+2)/(s^2+4s+13)) ?
The d/(ds)((s+2)/(s^2+4s+13)) is (-s^2-4s+5)/((s^2+4s+13)^2)
What is the first d/(ds)((s+2)/(s^2+4s+13)) ?
The first d/(ds)((s+2)/(s^2+4s+13)) is (-s^2-4s+5)/((s^2+4s+13)^2)