What is the derivative of x^4-\sqrt[3]{x} ?

The derivative of x^4-\sqrt[3]{x} is (-1+12x^{11/3})/(3x^{2/3)}

What is the first derivative of x^4-\sqrt[3]{x} ?

The first derivative of x^4-\sqrt[3]{x} is (-1+12x^{11/3})/(3x^{2/3)}

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\frac{- 1 + 12 x ^{\frac{11}{3}}}{3 x ^{\frac{2}{3}}}

Solution steps

Apply the Sum/Difference Rule:

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

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

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

Simplify

4 x^{3} - \frac{1}{3 x ^{\frac{2}{3}}} : \frac{- 1 + 12 x ^{\frac{11}{3}}}{3 x ^{\frac{2}{3}}}

= \frac{- 1 + 12 x ^{\frac{11}{3}}}{3 x ^{\frac{2}{3}}}

\int \frac{x + 3}{x ^{2} + 1} d x

\frac{d y}{d x} - \frac{2}{x} y = x^{2} - 1

y = 8 x - x^{2}, (1, 7)

x^{2} cos (x^{5})

\int \frac{x}{x ^{2} - 9} d x

What is the derivative of x^4-\sqrt[3]{x} ?
The derivative of x^4-\sqrt[3]{x} is (-1+12x^{11/3})/(3x^{2/3)}
What is the first derivative of x^4-\sqrt[3]{x} ?
The first derivative of x^4-\sqrt[3]{x} is (-1+12x^{11/3})/(3x^{2/3)}