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

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

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

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

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derivative of

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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