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

The derivative of \sqrt[3]{x}(x^2+4) is (x^2+4+6\sqrt[3]{x}x^{5/3})/(3x^{2/3)}

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

The first derivative of \sqrt[3]{x}(x^2+4) is (x^2+4+6\sqrt[3]{x}x^{5/3})/(3x^{2/3)}

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

\frac{x ^{2} + 4 + 6 3 x x ^{\frac{5}{3}}}{3 x ^{\frac{2}{3}}}

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 3 x, g = x^{2} + 4

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

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

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

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

Simplify

\frac{1}{3 x ^{\frac{2}{3}}} (x^{2} + 4) + 2 x 3 x : \frac{x ^{2} + 4 + 6 3 x x ^{\frac{5}{3}}}{3 x ^{\frac{2}{3}}}

= \frac{x ^{2} + 4 + 6 3 x x ^{\frac{5}{3}}}{3 x ^{\frac{2}{3}}}

\int e^{- v} d v

(2, 3), (4, - 4)

x^{2} - 2 x + 4

x \cdot e^{- x + 1}, 1

\int sin^{3} (θ) cos^{4} (θ) d θ

What is the derivative of \sqrt[3]{x}(x^2+4) ?
The derivative of \sqrt[3]{x}(x^2+4) is (x^2+4+6\sqrt[3]{x}x^{5/3})/(3x^{2/3)}
What is the first derivative of \sqrt[3]{x}(x^2+4) ?
The first derivative of \sqrt[3]{x}(x^2+4) is (x^2+4+6\sqrt[3]{x}x^{5/3})/(3x^{2/3)}