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

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

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

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

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

Solution steps

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (9 arctan (x))

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

x \to 0 lim (6 - \frac{1}{x ^{3}})

x \to 6 lim (- x^{2} + 3 x - 6)

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