# How do you find the first and second derivatives of (2x)/(sqrt(5x^2 -2x + 1)) using the quotient rule?

Oct 27, 2015

The first derivative is $f ' \left(x\right) = \frac{2 - 2 x}{5 {x}^{2} - 2 x + 1} ^ \left(\frac{3}{2}\right)$ and the second derivative is $f ' ' \left(x\right) = \frac{20 {x}^{2} - 32 x + 4}{5 {x}^{2} - 2 x + 1} ^ \left(\frac{5}{2}\right)$.

#### Explanation:

The quotient rule says $\frac{d}{\mathrm{dx}} \left(g \frac{x}{h \left(x\right)}\right) = \frac{h \left(x\right) \cdot g ' \left(x\right) - g \left(x\right) \cdot h ' \left(x\right)}{{\left(h \left(x\right)\right)}^{2}}$.

For $f \left(x\right) = \frac{2 x}{\sqrt{5 {x}^{2} - 2 x + 1}}$, we get

$f ' \left(x\right) = \frac{{\left(5 {x}^{2} - 2 x + 1\right)}^{\frac{1}{2}} \cdot 2 - \left(2 x\right) \cdot \frac{1}{2} \cdot {\left(5 {x}^{2} - 2 x + 1\right)}^{- \frac{1}{2}} \cdot \left(10 x - 2\right)}{5 {x}^{2} - 2 x + 1}$

$= \frac{10 {x}^{2} - 4 x + 2 - \left(10 {x}^{2} - 2 x\right)}{5 {x}^{2} - 2 x + 1} ^ \left(\frac{3}{2}\right) = \frac{2 - 2 x}{5 {x}^{2} - 2 x + 1} ^ \left(\frac{3}{2}\right)$

and then

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

$= \frac{{\left(5 {x}^{2} - 2 x + 1\right)}^{\frac{1}{2}} \cdot \left(- 10 {x}^{2} + 4 x - 2 - 3 \left(1 - x\right) \left(10 x - 2\right)\right)}{{\left(5 {x}^{2} - 2 x + 1\right)}^{3}}$

$= \frac{- 10 {x}^{2} + 4 x - 2 + 30 {x}^{2} - 36 x + 6}{{\left(5 {x}^{2} - 2 x + 1\right)}^{\frac{5}{2}}}$

$= \frac{20 {x}^{2} - 32 x + 4}{5 {x}^{2} - 2 x + 1} ^ \left(\frac{5}{2}\right)$.