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

Jun 19, 2018

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

#### Explanation:

The derivative of a quotient is

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

Here, we have

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

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

Therefore,

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

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

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

Jun 19, 2018

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

#### Explanation:

After the Quotient rule we get

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

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