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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int (x - 1) d x

t^{2} y^{^{''}} + 5 t y^{^{'}} - 12 y = 0

\frac{1}{( 1 - x ^{2} ) ^{2}}

10 x - 6 y x^{2} + 1 \frac{d y}{d x} = 0

\frac{d r}{d θ} = \frac{r ^{2} + 5 r θ ^{4}}{θ ^{5}}

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