# How do you find the first and second derivatives of f(x)=(x)/(x^2+1) using the quotient rule?

May 25, 2018

Below

#### Explanation:

$f \left(x\right) = \frac{x}{{x}^{2} + 1}$

$f ' \left(x\right) = \frac{\left({x}^{2} + 1\right) \left(1\right) - x \left(2 x\right)}{{x}^{2} + 1} ^ 2$

$f ' \left(x\right) = \frac{{x}^{2} + 1 - 2 {x}^{2}}{{x}^{2} + 1} ^ 2$

$f ' \left(x\right) = \frac{1 - {x}^{2}}{{x}^{2} + 1} ^ 2$

$f ' ' \left(x\right) = \frac{{\left({x}^{2} + 1\right)}^{2} \left(- 2 x\right) - \left(1 - {x}^{2}\right) \cdot \left({x}^{2} + 1\right) \left(2 x\right)}{{x}^{2} + 1} ^ 4$

$f ' ' \left(x\right) = \frac{- 2 x \left({x}^{2} + 1\right) - 2 x \left(1 - {x}^{2}\right)}{{x}^{2} + 1} ^ 3$

$f ' ' \left(x\right) = \frac{- 2 {x}^{3} - 2 x - 2 x + 2 {x}^{3}}{{x}^{2} + 1} ^ 3$

$f ' ' \left(x\right) = \frac{- 4 x}{{x}^{2} + 1} ^ 3$

The quotient rule is given by:

$f \left(x\right) = \frac{u}{v}$

$f ' \left(x\right) = \frac{v u ' - u v '}{v} ^ 2$

May 25, 2018

Please see the explanation below

#### Explanation:

The quotient rule is

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{{v}^{2}}$

Here,

$u = x$, $\implies$, $u ' = 1$

$v = {x}^{2} + 1$, $\implies$, $v ' = 2 x$

Therefore, the first derivative is

$f ' \left(x\right) = \frac{1 \left({x}^{2} + 1\right) - x \left(2 x\right)}{{x}^{2} + 1} ^ 2$

$= \frac{1 - {x}^{2}}{{x}^{2} + 1} ^ 2$

For the second derivative,

$u = \left(1 - {x}^{2}\right)$, $\implies$, $u ' = - 2 x$

$v = {\left({x}^{2} + 1\right)}^{2}$, $\implies$, $v ' = 4 x \left({x}^{2} + 1\right)$

Therefore, the second derivative is

$f ' ' \left(x\right) = \frac{- 2 x {\left({x}^{2} + 1\right)}^{2} - 4 x \left(1 - {x}^{2}\right) \left({x}^{2} + 1\right)}{{x}^{2} + 1} ^ 4$

$= \frac{\left({x}^{2} + 1\right) \left(- 2 x \left({x}^{2} + 1\right) - 4 x \left(1 - {x}^{2}\right)\right)}{{x}^{2} + 1} ^ 4$

$= \frac{2 {x}^{3} - 6 x}{{x}^{2} + 1} ^ 3$

$= \frac{2 x \left({x}^{2} - 3\right)}{{x}^{2} + 1} ^ 3$