What is the d/(ds)(4^{-3}(2/(s^7)+3s)^4) ?

The d/(ds)(4^{-3}(2/(s^7)+3s)^4) is ((-14+3s^8)(2+3s^8)^3)/(16s^{29)}

What is the first d/(ds)(4^{-3}(2/(s^7)+3s)^4) ?

The first d/(ds)(4^{-3}(2/(s^7)+3s)^4) is ((-14+3s^8)(2+3s^8)^3)/(16s^{29)}

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\frac{d}{d s} (4^{- 3} (\frac{2}{s ^{7}} + 3 s)^{4})

\frac{( - 14 + 3 s ^{8} ) ( 2 + 3 s ^{8} ) ^{3}}{16 s ^{29}}

Solution steps

\frac{d}{d s} (4^{- 3} (\frac{2}{s ^{7}} + 3 s)^{4})

Take the constant out:

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

= 4^{- 3} \frac{d}{d s} ((\frac{2}{s ^{7}} + 3 s)^{4})

Apply the chain rule:

4 (\frac{2}{s ^{7}} + 3 s)^{3} \frac{d}{d s} (\frac{2}{s ^{7}} + 3 s)

= 4 (\frac{2}{s ^{7}} + 3 s)^{3} \frac{d}{d s} (\frac{2}{s ^{7}} + 3 s)

\frac{d}{d s} (\frac{2}{s ^{7}} + 3 s) = - \frac{14}{s ^{8}} + 3

= 4^{- 3} \cdot 4 (\frac{2}{s ^{7}} + 3 s)^{3} (- \frac{14}{s ^{8}} + 3)

Simplify

4^{- 3} \cdot 4 (\frac{2}{s ^{7}} + 3 s)^{3} (- \frac{14}{s ^{8}} + 3) : \frac{( - 14 + 3 s ^{8} ) ( 2 + 3 s ^{8} ) ^{3}}{16 s ^{29}}

= \frac{( - 14 + 3 s ^{8} ) ( 2 + 3 s ^{8} ) ^{3}}{16 s ^{29}}

\int x^{3} ln ∣ 1 + x ∣ d x

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

f (x) = \frac{x ^{3} + 2 x ^{2} + 3 x + 4}{x}

\frac{d}{d x} (9 - x)

4 π x^{2}

What is the d/(ds)(4^{-3}(2/(s^7)+3s)^4) ?
The d/(ds)(4^{-3}(2/(s^7)+3s)^4) is ((-14+3s^8)(2+3s^8)^3)/(16s^{29)}
What is the first d/(ds)(4^{-3}(2/(s^7)+3s)^4) ?
The first d/(ds)(4^{-3}(2/(s^7)+3s)^4) is ((-14+3s^8)(2+3s^8)^3)/(16s^{29)}