# What's the derivative of f(x) = (arctan (x/2)) + ((5x-1)/(2(x^2) + 4))?

Sep 3, 2016

$f ' \left(x\right) = \frac{56 + 8 x + 6 {x}^{2} + 2 {x}^{3} - {x}^{4}}{2 \left({x}^{2} + 4\right) {\left({x}^{2} + 2\right)}^{2}}$.

#### Explanation:

Let $f \left(x\right) = g \left(x\right) + h \left(x\right) \text{, so that, } f ' \left(x\right) = g ' \left(x\right) + h ' \left(x\right) \ldots \ldots \ldots \left(\star\right)$,

where, $g \left(x\right) = a r c \tan \left(\frac{x}{2}\right) , \mathmr{and} , h \left(x\right) = \left\{\frac{5 x - 1}{2 {x}^{2} + 4}\right\}$.

$g \left(x\right) = a r c \tan \left(\frac{x}{2}\right) \Rightarrow g ' \left(x\right) = \left(\frac{1}{1 + {\left(\frac{x}{2}\right)}^{2}}\right) \cdot \frac{d}{\mathrm{dx}} \left(\frac{x}{2}\right)$

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

$h \left(x\right) = \frac{5 x - 1}{2 {x}^{2} + 4}$

$\therefore \ln h \left(x\right) = \ln \left(5 x - 1\right) - \ln \left(2 {x}^{2} + 4\right)$

$\therefore \frac{d}{\mathrm{dx}} \ln h \left(x\right) = \frac{d}{\mathrm{dx}} \ln \left(5 x - 1\right) - \frac{d}{\mathrm{dx}} \ln \left(2 {x}^{2} + 4\right)$

$\therefore \frac{1}{h \left(x\right)} \frac{d}{\mathrm{dx}} h \left(x\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \text{[by chain rule]}$

$= \frac{1}{5 x - 1} \frac{d}{\mathrm{dx}} \left(5 x - 1\right) - \frac{1}{2 {x}^{2} + 4} \cdot \frac{d}{\mathrm{dx}} \left(2 {x}^{2} + 4\right)$

$= \frac{5}{5 x - 1} - \frac{4 x}{2 {x}^{2} + 4}$

$\therefore h ' \left(x\right) = h \left(x\right) \left[\frac{5 \left(2 {x}^{2} + 4\right) - 4 x \left(5 x - 1\right)}{\left(5 x - 1\right) \left(2 {x}^{2} + 4\right)}\right]$

$\therefore h ' \left(x\right) = \frac{\cancel{\left(5 x - 1\right)}}{2 {x}^{2} + 4} \left\{\frac{20 + 4 x - 10 {x}^{2}}{\cancel{\left(5 x - 1\right)} \left(2 {x}^{2} + 4\right)}\right\}$

$\therefore h ' \left(x\right) = \frac{10 + 2 x - 5 {x}^{2}}{2 {\left({x}^{2} + 2\right)}^{2}} \ldots \ldots \ldots \ldots . \left(2\right)$

Using $\left(1\right) , \left(2\right) \text{ in } \left(\star\right)$,

$f ' \left(x\right) = \frac{2}{{x}^{2} + 4} + \frac{10 + 2 x - 5 {x}^{2}}{2 {\left({x}^{2} + 2\right)}^{2}}$

$= \frac{4 {\left({x}^{2} + 2\right)}^{2} + \left(4 + {x}^{2}\right) \left(10 + 2 x - 5 {x}^{2}\right)}{2 \left({x}^{2} + 4\right) {\left({x}^{2} + 2\right)}^{2}}$

$= \frac{4 {x}^{4} + 16 {x}^{2} + 16 + 40 + 8 x - 20 {x}^{2} + 10 {x}^{2} + 2 {x}^{3} - 5 {x}^{4}}{2 \left({x}^{2} + 4\right) {\left({x}^{2} + 2\right)}^{2}}$

$f ' \left(x\right) = \frac{56 + 8 x + 6 {x}^{2} + 2 {x}^{3} - {x}^{4}}{2 \left({x}^{2} + 4\right) {\left({x}^{2} + 2\right)}^{2}}$.