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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} ((9 x + 4) x^{2})

(40 q^{2} + q + 500)^{^{'}}

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

y = e^{x} + 1

x \to 0 lim (e^{- x} x^{2})

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