What is the d/(ds)((s^2-4s+20)/((s+2)(s+4))) ?

The d/(ds)((s^2-4s+20)/((s+2)(s+4))) is (10s^2-24s-152)/((s+2)^2(s+4)^2)

What is the first d/(ds)((s^2-4s+20)/((s+2)(s+4))) ?

The first d/(ds)((s^2-4s+20)/((s+2)(s+4))) is (10s^2-24s-152)/((s+2)^2(s+4)^2)

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\frac{d}{d s} (\frac{s ^{2} - 4 s + 20}{( s + 2 ) ( s + 4 )})

\frac{10 s ^{2} - 24 s - 152}{( s + 2 ) ^{2} ( s + 4 ) ^{2}}

Solution steps

\frac{d}{d s} (\frac{s ^{2} - 4 s + 20}{( s + 2 ) ( s + 4 )})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d s} ( s ^{2} - 4 s + 20 ) ( s + 2 ) ( s + 4 ) - \frac{d}{d s} ( ( s + 2 ) ( s + 4 ) ) ( s ^{2} - 4 s + 20 )}{( ( s + 2 ) ( s + 4 ) ) ^{2}}

\frac{d}{d s} (s^{2} - 4 s + 20) = 2 s - 4

\frac{d}{d s} ((s + 2) (s + 4)) = 2 s + 6

= \frac{( 2 s - 4 ) ( s + 2 ) ( s + 4 ) - ( 2 s + 6 ) ( s ^{2} - 4 s + 20 )}{( ( s + 2 ) ( s + 4 ) ) ^{2}}

Simplify

\frac{( 2 s - 4 ) ( s + 2 ) ( s + 4 ) - ( 2 s + 6 ) ( s ^{2} - 4 s + 20 )}{( ( s + 2 ) ( s + 4 ) ) ^{2}} : \frac{10 s ^{2} - 24 s - 152}{( s + 2 ) ^{2} ( s + 4 ) ^{2}}

= \frac{10 s ^{2} - 24 s - 152}{( s + 2 ) ^{2} ( s + 4 ) ^{2}}

\int e^{t a n (6 x)} sec^{2} (6 x) d x

y^{^{'}} = 6 x^{2} y

\frac{d y}{d x}, 4 x = y^{2}

\frac{e ^{x}}{3 x ^{2} + 7}

\int 5 + x d x

What is the d/(ds)((s^2-4s+20)/((s+2)(s+4))) ?
The d/(ds)((s^2-4s+20)/((s+2)(s+4))) is (10s^2-24s-152)/((s+2)^2(s+4)^2)
What is the first d/(ds)((s^2-4s+20)/((s+2)(s+4))) ?
The first d/(ds)((s^2-4s+20)/((s+2)(s+4))) is (10s^2-24s-152)/((s+2)^2(s+4)^2)