What is the derivative of (x^3-9x^{-1/3}) ?

The derivative of (x^3-9x^{-1/3}) is -(x^2-3)/((x^3-9x)^{4/3)}

What is the first derivative of (x^3-9x^{-1/3}) ?

The first derivative of (x^3-9x^{-1/3}) is -(x^2-3)/((x^3-9x)^{4/3)}

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\frac{d}{d x} ((x^{3} - 9 x)^{- \frac{1}{3}})

- \frac{x ^{2} - 3}{( x ^{3} - 9 x ) ^{\frac{4}{3}}}

Solution steps

\frac{d}{d x} ((x^{3} - 9 x)^{- \frac{1}{3}})

Apply the chain rule:

- \frac{1}{3 ( x ^{3} - 9 x ) ^{\frac{4}{3}}} \frac{d}{d x} (x^{3} - 9 x)

= - \frac{1}{3 ( x ^{3} - 9 x ) ^{\frac{4}{3}}} \frac{d}{d x} (x^{3} - 9 x)

\frac{d}{d x} (x^{3} - 9 x) = 3 x^{2} - 9

= - \frac{1}{3 ( x ^{3} - 9 x ) ^{\frac{4}{3}}} (3 x^{2} - 9)

Simplify

- \frac{1}{3 ( x ^{3} - 9 x ) ^{\frac{4}{3}}} (3 x^{2} - 9) : - \frac{x ^{2} - 3}{( x ^{3} - 9 x ) ^{\frac{4}{3}}}

= - \frac{x ^{2} - 3}{( x ^{3} - 9 x ) ^{\frac{4}{3}}}

\frac{d}{d x} (tan (cot (ln (x))) + sec (e^{x}))

x \to 0 + lim (\frac{sin ( x )}{x})

\frac{\partial}{\partial x} (x^{3} y z^{2} + 4 yz)

\int 4 x^{5} e^{x^{6}} d x

\frac{\partial}{\partial y} (6 x + 10 y)

What is the derivative of (x^3-9x^{-1/3}) ?
The derivative of (x^3-9x^{-1/3}) is -(x^2-3)/((x^3-9x)^{4/3)}
What is the first derivative of (x^3-9x^{-1/3}) ?
The first derivative of (x^3-9x^{-1/3}) is -(x^2-3)/((x^3-9x)^{4/3)}