What is the derivative of \sqrt[3]{8+4x-x^3} ?

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

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

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

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

Solution steps

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 0 - lim (x^{2} csc (x))

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

x \to 0 lim (sin (2 x) csc (3 x))

\frac{d p}{d x}, x^{2} + 3 p x + p^{2} = 79

\frac{d}{d x} (- \frac{9}{( x + 1 ) ^{2}} + 1)

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