What is the derivative of 2/(\sqrt[3]{4x+1}) ?

The derivative of 2/(\sqrt[3]{4x+1}) is -8/(3(4x+1)^{4/3)}

What is the first derivative of 2/(\sqrt[3]{4x+1}) ?

The first derivative of 2/(\sqrt[3]{4x+1}) is -8/(3(4x+1)^{4/3)}

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- \frac{8}{3 ( 4 x + 1 ) ^{\frac{4}{3}}}

Solution steps

Take the constant out:

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

\frac{d}{d x} (4 x + 1) = 4

= 2 (- \frac{1}{3 ( 4 x + 1 ) ^{\frac{4}{3}}} \cdot 4)

Simplify

2 (- \frac{1}{3 ( 4 x + 1 ) ^{\frac{4}{3}}} \cdot 4) : - \frac{8}{3 ( 4 x + 1 ) ^{\frac{4}{3}}}

= - \frac{8}{3 ( 4 x + 1 ) ^{\frac{4}{3}}}

\frac{d}{d x} (4 x sin (π x))

f (x) = \frac{x ^{2}}{x + 8}, at x = 2

\int 10 x ln (9 x) d x

\int_{- π}^{0} sin (x) d x

\frac{d}{d x} (\frac{( x ^{2} )}{x - 1})

What is the derivative of 2/(\sqrt[3]{4x+1}) ?
The derivative of 2/(\sqrt[3]{4x+1}) is -8/(3(4x+1)^{4/3)}
What is the first derivative of 2/(\sqrt[3]{4x+1}) ?
The first derivative of 2/(\sqrt[3]{4x+1}) is -8/(3(4x+1)^{4/3)}