What is the derivative of (x/(x^3+1)^6) ?

The derivative of (x/(x^3+1)^6) is (6x^5(-2x^3+1))/((x^3+1)^7)

What is the first derivative of (x/(x^3+1)^6) ?

The first derivative of (x/(x^3+1)^6) is (6x^5(-2x^3+1))/((x^3+1)^7)

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\frac{d}{d x} ((\frac{x}{x ^{3} + 1})^{6})

\frac{6 x ^{5} ( - 2 x ^{3} + 1 )}{( x ^{3} + 1 ) ^{7}}

Solution steps

\frac{d}{d x} ((\frac{x}{x ^{3} + 1})^{6})

Apply the chain rule:

6 (\frac{x}{x ^{3} + 1})^{5} \frac{d}{d x} (\frac{x}{x ^{3} + 1})

= 6 (\frac{x}{x ^{3} + 1})^{5} \frac{d}{d x} (\frac{x}{x ^{3} + 1})

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

= 6 (\frac{x}{x ^{3} + 1})^{5} \frac{- 2 x ^{3} + 1}{( x ^{3} + 1 ) ^{2}}

Simplify

6 (\frac{x}{x ^{3} + 1})^{5} \frac{- 2 x ^{3} + 1}{( x ^{3} + 1 ) ^{2}} : \frac{6 x ^{5} ( - 2 x ^{3} + 1 )}{( x ^{3} + 1 ) ^{7}}

= \frac{6 x ^{5} ( - 2 x ^{3} + 1 )}{( x ^{3} + 1 ) ^{7}}

9 tan^{2} (6 x)

\frac{d}{d x} ((x^{2} + 15)^{2})

e^{2 t} cos (5 t)

y = 3 x

x^{2} + y + 4 = 0, y = - 8

What is the derivative of (x/(x^3+1)^6) ?
The derivative of (x/(x^3+1)^6) is (6x^5(-2x^3+1))/((x^3+1)^7)
What is the first derivative of (x/(x^3+1)^6) ?
The first derivative of (x/(x^3+1)^6) is (6x^5(-2x^3+1))/((x^3+1)^7)