# How do you differentiate f(x)=(x-2)/(x^4+1)?

Sep 6, 2017

$f ' \left(x\right) = \frac{8 {x}^{3} - 3 {x}^{4} + 1}{{x}^{4} + 1} ^ 2$

#### Explanation:

$\text{differentiate using the "color(blue)"quotient rule}$

$\text{given "f(x)=(g(x))/(h(x))" then}$

$f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2 \leftarrow \text{ quotient rule}$

$g \left(x\right) = x - 2 \Rightarrow g ' \left(x\right) = 1$

$h \left(x\right) = {x}^{4} + 1 \Rightarrow h ' \left(x\right) = 4 {x}^{3}$

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

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = \frac{8 {x}^{3} - 3 {x}^{4} + 1}{{x}^{4} + 1} ^ 2$

Sep 6, 2017

$f ' \left(x\right) = \frac{8 {x}^{3} - 3 {x}^{4} + 1}{{x}^{4} + 1} ^ 2$

#### Explanation:

We could simply use the u/v rule of differentiation, which is

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

where,

$u$ and $v$ are diffrentiable functions and $u '$ and $v '$ are their derivatives.

In this question,

$u = \left(x - 2\right)$ and, $v = \left({x}^{4} + 1\right)$

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

On differentiating the function we have,

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

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

$\implies f ' \left(x\right) = \frac{8 {x}^{3} - 3 {x}^{4} + 1}{{x}^{4} + 1} ^ 2$