What is the derivative of \sqrt[3]{x^3-2x^2+8} ?

The derivative of \sqrt[3]{x^3-2x^2+8} is (3x^2-4x)/(3(x^3-2x^2+8)^{2/3)}

What is the first derivative of \sqrt[3]{x^3-2x^2+8} ?

The first derivative of \sqrt[3]{x^3-2x^2+8} is (3x^2-4x)/(3(x^3-2x^2+8)^{2/3)}

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

Solution steps

Apply the chain rule:

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

= \frac{1}{3 ( x ^{3} - 2 x ^{2} + 8 ) ^{\frac{2}{3}}} \frac{d}{d x} (x^{3} - 2 x^{2} + 8)

\frac{d}{d x} (x^{3} - 2 x^{2} + 8) = 3 x^{2} - 4 x

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

Simplify

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

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

cos (sin (tan (3 x)))

\frac{2}{5 ( s + 1 ) ^{2} - 80}

x \to 2 lim (\frac{1 - x ^{2}}{1 - x})

x \to - 3 + lim (x x + 3)

y = ∣ x ∣, y = x^{2} - 2

What is the derivative of \sqrt[3]{x^3-2x^2+8} ?
The derivative of \sqrt[3]{x^3-2x^2+8} is (3x^2-4x)/(3(x^3-2x^2+8)^{2/3)}
What is the first derivative of \sqrt[3]{x^3-2x^2+8} ?
The first derivative of \sqrt[3]{x^3-2x^2+8} is (3x^2-4x)/(3(x^3-2x^2+8)^{2/3)}