What is the derivative of f(x) = sqrt[ (3 x + 1) / (5 x^2 + 1) ]?

Mar 27, 2018

Use chain rule: if $f \left(x\right) = h \left(g \left(x\right)\right)$ the derivative is$f ' \left(x\right) = h ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$
and quotient rule:
${\left(\frac{f}{g}\right)}^{'} = \left\{\setminus \frac{f ' g - f g '}{{g}^{2}}\right\}$ where $g \ne 0$

Explanation:

$f ' \left(x\right) = {\left({\left(\frac{3 x + 1}{5 {x}^{2} + 1}\right)}^{\frac{1}{2}}\right)}^{'}$
$= \frac{1}{2} \cdot {\left(\frac{3 x + 1}{5 {x}^{2} + 1}\right)}^{- \frac{1}{2}} \cdot {\left(\frac{3 x + 1}{5 {x}^{2} + 1}\right)}^{'}$

Lets calculate ${\left(\frac{3 x + 1}{5 {x}^{2} + 1}\right)}^{'}$ separately

${\left(\frac{3 x + 1}{5 {x}^{2} + 1}\right)}^{'} =$
$= \frac{{\left(3 x + 1\right)}^{'} \cdot \left(5 {x}^{2} + 1\right) - \left(3 x + 1\right) \cdot {\left(5 {x}^{2} + 1\right)}^{'}}{{\left(5 {x}^{2} + 1\right)}^{2}} =$

$= \frac{3 \cdot \left(5 {x}^{2} + 1\right) - \left(3 x + 1\right) \cdot 10 x}{{\left(5 {x}^{2} + 1\right)}^{2}}$
$= \frac{15 {x}^{2} + 3 - 30 {x}^{2} - 10 x}{{\left(5 {x}^{2} + 1\right)}^{2}} =$
$= \frac{- 15 {x}^{2} - 10 x + 3}{{\left(5 {x}^{2} + 1\right)}^{2}}$

The result is:
f'(x)=(-15x^2-10x+3)/(2(5 x^2 + 1)^2 ((3 x + 1)/ (5 x^2 + 1))^(1/2)