What is the derivative of \sqrt[3]{6x^2-x^3} ?

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

What is the first derivative of \sqrt[3]{6x^2-x^3} ?

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

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\frac{x ( - x + 4 )}{( 6 x ^{2} - x ^{3} ) ^{\frac{2}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( 6 x ^{2} - x ^{3} ) ^{\frac{2}{3}}} \frac{d}{d x} (6 x^{2} - x^{3})

= \frac{1}{3 ( 6 x ^{2} - x ^{3} ) ^{\frac{2}{3}}} \frac{d}{d x} (6 x^{2} - x^{3})

\frac{d}{d x} (6 x^{2} - x^{3}) = 12 x - 3 x^{2}

= \frac{1}{3 ( 6 x ^{2} - x ^{3} ) ^{\frac{2}{3}}} (12 x - 3 x^{2})

Simplify

\frac{1}{3 ( 6 x ^{2} - x ^{3} ) ^{\frac{2}{3}}} (12 x - 3 x^{2}) : \frac{x ( - x + 4 )}{( 6 x ^{2} - x ^{3} ) ^{\frac{2}{3}}}

= \frac{x ( - x + 4 )}{( 6 x ^{2} - x ^{3} ) ^{\frac{2}{3}}}

x \to 0 lim (\frac{∣ x ∣}{x ^{2} + x})

\int tan (6 x se) c^{2} 6 x d x

\frac{9 e ^{2 w} + e ^{w}}{e ^{w}}

\frac{d y}{d x} = x^{4} y^{- 3}

y^{^{''}} + 2 y^{^{'}} - 3 y = - 2 e^{3 t}

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