# How do you find f'(x) and f''(x) given f(x)=x^2/(1+2x)?

Aug 1, 2017

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

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

#### Explanation:

We're asked to find the first and second derivative of

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

First Derivative:

Use the quotient rule, which states

$\frac{d}{\mathrm{dx}} \left[\frac{u}{v}\right] = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{{v}^{2}}$

where

• $u = {x}^{2}$

• $v = 2 x + 1$:

$= \frac{\left(2 x + 1\right) \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] - {x}^{2} \frac{d}{\mathrm{dx}} \left[2 x + 1\right]}{{\left(2 x + 1\right)}^{2}}$

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

= color(red)((2x^2+2x)/((2x+1)^2)

Second derivative:

Use quotient rule again, this time

• $u = 2 {x}^{2} + 2 x$

• $v = {\left(2 x + 1\right)}^{2}$:

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

Use the chain rule on the ${\left(2 x + 1\right)}^{2}$ term:

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

Simplifying yields

= color(blue)(2/((2x+1)^3)