What is the derivative of 1/(x^3-4x) ?

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

What is the first derivative of 1/(x^3-4x) ?

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

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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