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

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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

(- x^{2})^{^{'}}

\frac{d}{d x} (x e^{x + 1})

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

\int_{0}^{2} 2 t^{3} d t

\frac{d}{d x} (x + y) = 2

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