# How do you find the derivative of x/(1+x^2)?

Jun 6, 2015

You can do this two ways.

a)
Quotient Rule:
d/(dx)[(f(x))/(g(x))] = (g(x)(df(x))/(dx) - f(x)(dg(x))/(dx))/((g(x))^2

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

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

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

$\frac{d}{\mathrm{dx}} \left[f \left(x\right) \cdot g \left(x\right)\right] = f \left(x\right) \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} + g \left(x\right) \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}}$

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

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

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

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

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