What is the derivative of (s^2+1/s)^{3/2} ?

The derivative of (s^2+1/s)^{3/2} is (3(2s^3-1)(s^3+1)^{1/2})/(2s^{5/2)}

What is the first derivative of (s^2+1/s)^{3/2} ?

The first derivative of (s^2+1/s)^{3/2} is (3(2s^3-1)(s^3+1)^{1/2})/(2s^{5/2)}

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derivative of

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

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

Solution steps

\frac{d}{d s} ((s^{2} + \frac{1}{s})^{\frac{3}{2}})

Apply the chain rule:

\frac{3 ( s ^{2} + \frac{1}{s} ) ^{\frac{1}{2}}}{2} \frac{d}{d s} (s^{2} + \frac{1}{s})

= \frac{3 ( s ^{2} + \frac{1}{s} ) ^{\frac{1}{2}}}{2} \frac{d}{d s} (s^{2} + \frac{1}{s})

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

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

Simplify

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

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

x \to 1 + lim (\frac{x + 8}{x - 1})

\int - \frac{1}{t - 2} d t

y^{^{'}} + 4 x y^{2} = 0

\frac{d y}{1 + 4 y ^{2}} = - \frac{x d x}{1 + x ^{4}}

\int \frac{5 x - 1}{( x + 1 ) ( x - 2 )} d x

What is the derivative of (s^2+1/s)^{3/2} ?
The derivative of (s^2+1/s)^{3/2} is (3(2s^3-1)(s^3+1)^{1/2})/(2s^{5/2)}
What is the first derivative of (s^2+1/s)^{3/2} ?
The first derivative of (s^2+1/s)^{3/2} is (3(2s^3-1)(s^3+1)^{1/2})/(2s^{5/2)}