What is the d/(ds)((6s^3+28s^2+38s+15)/((s+1)^2)) ?

The d/(ds)((6s^3+28s^2+38s+15)/((s+1)^2)) is (2(3s^3+9s^2+9s+4))/((s+1)^3)

What is the first d/(ds)((6s^3+28s^2+38s+15)/((s+1)^2)) ?

The first d/(ds)((6s^3+28s^2+38s+15)/((s+1)^2)) is (2(3s^3+9s^2+9s+4))/((s+1)^3)

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\frac{d}{d s} (\frac{6 s ^{3} + 28 s ^{2} + 38 s + 15}{( s + 1 ) ^{2}})

\frac{2 ( 3 s ^{3} + 9 s ^{2} + 9 s + 4 )}{( s + 1 ) ^{3}}

Solution steps

\frac{d}{d s} (\frac{6 s ^{3} + 28 s ^{2} + 38 s + 15}{( s + 1 ) ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d s} ( 6 s ^{3} + 28 s ^{2} + 38 s + 15 ) ( s + 1 ) ^{2} - \frac{d}{d s} ( ( s + 1 ) ^{2} ) ( 6 s ^{3} + 28 s ^{2} + 38 s + 15 )}{( ( s + 1 ) ^{2} ) ^{2}}

\frac{d}{d s} (6 s^{3} + 28 s^{2} + 38 s + 15) = 18 s^{2} + 56 s + 38

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

= \frac{( 18 s ^{2} + 56 s + 38 ) ( s + 1 ) ^{2} - 2 ( s + 1 ) ( 6 s ^{3} + 28 s ^{2} + 38 s + 15 )}{( ( s + 1 ) ^{2} ) ^{2}}

Simplify

\frac{( 18 s ^{2} + 56 s + 38 ) ( s + 1 ) ^{2} - 2 ( s + 1 ) ( 6 s ^{3} + 28 s ^{2} + 38 s + 15 )}{( ( s + 1 ) ^{2} ) ^{2}} : \frac{2 ( 3 s ^{3} + 9 s ^{2} + 9 s + 4 )}{( s + 1 ) ^{3}}

= \frac{2 ( 3 s ^{3} + 9 s ^{2} + 9 s + 4 )}{( s + 1 ) ^{3}}

\frac{d}{d x} (2 x^{3} - 9 x^{2} + 12 x - 3)

f (x) = \frac{1}{x ^{8}}

\frac{d}{d x} (ln (ln (15 x)))

t sin (π t)

\int_{3}^{4.5} \frac{x}{8} d x

What is the d/(ds)((6s^3+28s^2+38s+15)/((s+1)^2)) ?
The d/(ds)((6s^3+28s^2+38s+15)/((s+1)^2)) is (2(3s^3+9s^2+9s+4))/((s+1)^3)
What is the first d/(ds)((6s^3+28s^2+38s+15)/((s+1)^2)) ?
The first d/(ds)((6s^3+28s^2+38s+15)/((s+1)^2)) is (2(3s^3+9s^2+9s+4))/((s+1)^3)