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

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

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

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

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derivative of

y = \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}}

3^{x} \cdot x^{3}

2^{c o s^{2} (x)}

4 \frac{t ^{5}}{5} - \frac{t ^{3}}{6}

\frac{d}{d q} (\frac{5 + 1}{( cot ( q ) )})

f (x) = 3 cos (x) - 4 sec (x) + 3

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