What is the d/(ds)(4/(s^2+16)) ?

The d/(ds)(4/(s^2+16)) is -(8s)/((s^2+16)^2)

What is the first d/(ds)(4/(s^2+16)) ?

The first d/(ds)(4/(s^2+16)) is -(8s)/((s^2+16)^2)

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\frac{d}{d s} (\frac{4}{s ^{2} + 16})

- \frac{8 s}{( s ^{2} + 16 ) ^{2}}

Solution steps

\frac{d}{d s} (\frac{4}{s ^{2} + 16})

Take the constant out:

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

= 4 \frac{d}{d s} (\frac{1}{s ^{2} + 16})

Apply exponent rule:

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

= 4 \frac{d}{d s} ((s^{2} + 16)^{- 1})

Apply the chain rule:

- \frac{1}{( s ^{2} + 16 ) ^{2}} \frac{d}{d s} (s^{2} + 16)

= - \frac{1}{( s ^{2} + 16 ) ^{2}} \frac{d}{d s} (s^{2} + 16)

\frac{d}{d s} (s^{2} + 16) = 2 s

= 4 (- \frac{1}{( s ^{2} + 16 ) ^{2}} \cdot 2 s)

Simplify

4 (- \frac{1}{( s ^{2} + 16 ) ^{2}} \cdot 2 s) : - \frac{8 s}{( s ^{2} + 16 ) ^{2}}

= - \frac{8 s}{( s ^{2} + 16 ) ^{2}}

\int \frac{1 - 2 x}{e ^{2 x}} d x

\frac{d}{d x} (\frac{1}{2} x + \frac{1}{4} sin (2 x))

\int x^{2} e^{x^{3} - 1} d x

\frac{\partial}{\partial x} (x ln (x y - 9))

\frac{4 x ^{2} + 2 x + 8}{x}

What is the d/(ds)(4/(s^2+16)) ?
The d/(ds)(4/(s^2+16)) is -(8s)/((s^2+16)^2)
What is the first d/(ds)(4/(s^2+16)) ?
The first d/(ds)(4/(s^2+16)) is -(8s)/((s^2+16)^2)