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

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

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

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

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derivative of

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}}

n = 0 \sum \infty \frac{1}{2} (\frac{1}{2})^{n}

\int \frac{2 x + 3}{x ^{2} + 1} d x

x \to 4 - lim (16 - x^{2})

\frac{d y}{d x}, 1 = (\frac{y}{x})

f (x) = 8 cos (t)

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