What is the derivative of x^{8/3}-20x^{2/3} ?

The derivative of x^{8/3}-20x^{2/3} is (8x^2-40)/(3x^{1/3)}

What is the first derivative of x^{8/3}-20x^{2/3} ?

The first derivative of x^{8/3}-20x^{2/3} is (8x^2-40)/(3x^{1/3)}

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\frac{d}{d x} (x^{\frac{8}{3}} - 20 x^{\frac{2}{3}})

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

Solution steps

\frac{d}{d x} (x^{\frac{8}{3}} - 20 x^{\frac{2}{3}})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (x^{\frac{8}{3}}) - \frac{d}{d x} (20 x^{\frac{2}{3}})

\frac{d}{d x} (x^{\frac{8}{3}}) = \frac{8 x ^{\frac{5}{3}}}{3}

\frac{d}{d x} (20 x^{\frac{2}{3}}) = \frac{40}{3 x ^{\frac{1}{3}}}

= \frac{8 x ^{\frac{5}{3}}}{3} - \frac{40}{3 x ^{\frac{1}{3}}}

Simplify

\frac{8 x ^{\frac{5}{3}}}{3} - \frac{40}{3 x ^{\frac{1}{3}}} : \frac{8 x ^{2} - 40}{3 x ^{\frac{1}{3}}}

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

\frac{d y}{d x} = y^{2} sin (x^{2}), y (- 2) = \frac{1}{3}

f (r) = π r^{2}

2 x x, at x = 1

6 x

\int (2 x y - tan (y)) d x

What is the derivative of x^{8/3}-20x^{2/3} ?
The derivative of x^{8/3}-20x^{2/3} is (8x^2-40)/(3x^{1/3)}
What is the first derivative of x^{8/3}-20x^{2/3} ?
The first derivative of x^{8/3}-20x^{2/3} is (8x^2-40)/(3x^{1/3)}