What is the derivative of f(x)=(x^3-4/x)^{-6} ?

The derivative of f(x)=(x^3-4/x)^{-6} is -(6x^5(3x^4+4))/((x^4-4)^7)

What is the first derivative of f(x)=(x^3-4/x)^{-6} ?

The first derivative of f(x)=(x^3-4/x)^{-6} is -(6x^5(3x^4+4))/((x^4-4)^7)

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derivative of

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

- \frac{6 x ^{5} ( 3 x ^{4} + 4 )}{( x ^{4} - 4 ) ^{7}}

Solution steps

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

Apply the chain rule:

- \frac{6}{( x ^{3} - \frac{4}{x} ) ^{7}} \frac{d}{d x} (x^{3} - \frac{4}{x})

= - \frac{6}{( x ^{3} - \frac{4}{x} ) ^{7}} \frac{d}{d x} (x^{3} - \frac{4}{x})

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

= - \frac{6}{( x ^{3} - \frac{4}{x} ) ^{7}} (3 x^{2} + \frac{4}{x ^{2}})

Simplify

- \frac{6}{( x ^{3} - \frac{4}{x} ) ^{7}} (3 x^{2} + \frac{4}{x ^{2}}) : - \frac{6 x ^{5} ( 3 x ^{4} + 4 )}{( x ^{4} - 4 ) ^{7}}

= - \frac{6 x ^{5} ( 3 x ^{4} + 4 )}{( x ^{4} - 4 ) ^{7}}

\frac{d}{d x} (arcsin (ln (k x + 9)))

\int 2 e^{- t} t d t

\int e^{- 10 r^{2}} d r

(x - 1) \cdot ln (x - 1), 2

9 - x^{2}, 0

What is the derivative of f(x)=(x^3-4/x)^{-6} ?
The derivative of f(x)=(x^3-4/x)^{-6} is -(6x^5(3x^4+4))/((x^4-4)^7)
What is the first derivative of f(x)=(x^3-4/x)^{-6} ?
The first derivative of f(x)=(x^3-4/x)^{-6} is -(6x^5(3x^4+4))/((x^4-4)^7)