What is the derivative of 1/(x^4+x^2+1) ?

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

What is the first derivative of 1/(x^4+x^2+1) ?

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

- 2 sin (2 t)

\int \frac{4}{3 + 4 x} d x

\int_{1}^{49} \frac{1}{2 x} d x

\frac{d}{d x} (\frac{x ^{2}}{3})

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