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

The d/(ds)((s-sqrt(s))/(s^2)) is (-2sqrt(s)+3)/(2s^{5/2)}

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

The first d/(ds)((s-sqrt(s))/(s^2)) is (-2sqrt(s)+3)/(2s^{5/2)}

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\frac{d}{d s} (\frac{s - s}{s ^{2}})

\frac{- 2 s + 3}{2 s ^{\frac{5}{2}}}

Solution steps

\frac{d}{d s} (\frac{s - s}{s ^{2}})

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( 1 - \frac{1}{2 s} ) s ^{2} - 2 s ( s - s )}{( s ^{2} ) ^{2}} : \frac{- 2 s + 3}{2 s ^{\frac{5}{2}}}

= \frac{- 2 s + 3}{2 s ^{\frac{5}{2}}}

3 - 2 x

\frac{d}{d x} ((2 x + 1)^{3} e^{- (1 - x)^{4}})

f (x) = \frac{1}{x + 6}

\frac{d}{d v} (\frac{1}{v})

2 y^{^{'}} - y = 4

What is the d/(ds)((s-sqrt(s))/(s^2)) ?
The d/(ds)((s-sqrt(s))/(s^2)) is (-2sqrt(s)+3)/(2s^{5/2)}
What is the first d/(ds)((s-sqrt(s))/(s^2)) ?
The first d/(ds)((s-sqrt(s))/(s^2)) is (-2sqrt(s)+3)/(2s^{5/2)}