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

The derivative of x^{2/3}(x-4^2) is (8x^2-40x+32)/(3x^{1/3)}

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

The first derivative of x^{2/3}(x-4^2) is (8x^2-40x+32)/(3x^{1/3)}

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

\frac{8 x ^{2} - 40 x + 32}{3 x ^{\frac{1}{3}}}

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

\frac{2}{3 x ^{\frac{1}{3}}} (x - 4)^{2} + 2 (x - 4) x^{\frac{2}{3}} : \frac{8 x ^{2} - 40 x + 32}{3 x ^{\frac{1}{3}}}

= \frac{8 x ^{2} - 40 x + 32}{3 x ^{\frac{1}{3}}}

\int_{0}^{1} x^{3} e^{- x^{2}} d x

\frac{d}{d t} (e^{8 t s i n (2 t)})

\frac{d}{d x} (10, 0 - 40 x - 0.02 x^{2})

x \to c lim (x^{2} + a)

\frac{1}{8 x}

What is the derivative of x^{2/3}(x-4^2) ?
The derivative of x^{2/3}(x-4^2) is (8x^2-40x+32)/(3x^{1/3)}
What is the first derivative of x^{2/3}(x-4^2) ?
The first derivative of x^{2/3}(x-4^2) is (8x^2-40x+32)/(3x^{1/3)}