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

The derivative of x^{2/3}(x^2-4) is (8x^2-8)/(3x^{1/3)}

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

The first derivative of x^{2/3}(x^2-4) is (8x^2-8)/(3x^{1/3)}

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

\frac{8 x ^{2} - 8}{3 x ^{\frac{1}{3}}}

Solution steps

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

Apply the Product Rule:

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

f = x^{\frac{2}{3}}, g = x^{2} - 4

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

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

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

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

Simplify

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

= \frac{8 x ^{2} - 8}{3 x ^{\frac{1}{3}}}

n = 1 \sum \infty \frac{8 ^{n}}{9 ^{n}}

\int_{- \infty}^{- 1} \frac{1}{x ^{3}} d x

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

f (x) = 6 e^{x}

h \to + 6 lim (\frac{4 h}{h - 6})

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