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

The d/(ds)(2/(s+sqrt(s^2+4))) is -1/2 (1-s/(sqrt(s^2+4)))

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

The first d/(ds)(2/(s+sqrt(s^2+4))) is -1/2 (1-s/(sqrt(s^2+4)))

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

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

Solution steps

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

Simplify

\frac{2}{s + s ^{2} + 4} : - \frac{s - s ^{2} + 4}{2}

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

Take the constant out:

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

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d s}{d s} = 1

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

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

\frac{\partial}{\partial x} (6 x y - 12 x)

(2 x + 4) d x + (3 y - 1) d y = 0

x \to 0 lim (\frac{1}{x + 3})

\int cos^{4} (2 x) sin (2 x) d x

\frac{d}{d x} (\frac{cos ( x )}{x ^{2}})

What is the d/(ds)(2/(s+sqrt(s^2+4))) ?
The d/(ds)(2/(s+sqrt(s^2+4))) is -1/2 (1-s/(sqrt(s^2+4)))
What is the first d/(ds)(2/(s+sqrt(s^2+4))) ?
The first d/(ds)(2/(s+sqrt(s^2+4))) is -1/2 (1-s/(sqrt(s^2+4)))