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

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

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

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

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

- \frac{1}{( s ^{2} + 4 ) ^{2}} \frac{d}{d s} (s^{2} + 4)

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

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

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

Simplify

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

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

\int_{1}^{2} x^{2} d x

\int_{0}^{2} \frac{x}{16 - x ^{2}} d x

\int \frac{5}{2 x + 7} d x

x \to 2 lim (\frac{x + 4}{3 x - 1})

\int \frac{1}{( 4 x ^{2} - 1 ) ^{2}} d x

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