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

May 8, 2017

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1 - {x}^{2}}{1 + {x}^{2}} ^ 2.$

#### Explanation:

Knowing that, $\sin 2 \theta = \frac{2 \tan \theta}{1 + {\tan}^{2} \theta} ,$ we put,

$x = \tan \theta$ in $u$, and find that,

$u = \tan \frac{\theta}{1 + {\tan}^{2} \theta} = \frac{1}{2} \sin 2 \theta .$

Note that, as the Range of $\tan$ fun. is $\mathbb{R} ,$ we can put $x = \tan \theta .$

Thus, $u$ is a fun. of $\theta$, and, $\theta$ of $x .$

Hence, by the Chain Rule, we have,

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\mathrm{du}}{d \left(\theta\right)} \cdot \frac{d \left(\theta\right)}{\mathrm{dx}} \ldots \ldots \ldots . \left(\star\right) .$

Now, $u = \frac{1}{2} \sin 2 \theta \Rightarrow \frac{\mathrm{du}}{d \left(\theta\right)} = \frac{1}{2} \left(\cos 2 \theta\right) \left(2\right) , i . e . ,$

$\frac{\mathrm{du}}{d \left(\theta\right)} = \cos 2 \theta \ldots \ldots \ldots \ldots \ldots . . \left(1\right) .$

$x = \tan \theta \Rightarrow \theta = a r c \tan x \Rightarrow \frac{d \left(\theta\right)}{\mathrm{dx}} = \frac{1}{1 + {x}^{2}} \ldots . \left(2\right) .$

Therefore, by $\left(\star\right) , \left(1\right) \mathmr{and} \left(2\right) ,$ we have,

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\cos 2 \theta}{1 + {x}^{2}} , \text{ and, finally, since,}$

$\cos 2 \theta = \frac{1 - {\tan}^{2} \theta}{1 + {\tan}^{2} \theta} = \frac{1 - {x}^{2}}{1 + {x}^{2}} ,$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1 - {x}^{2}}{1 + {x}^{2}} ^ 2.$

Enjoy Maths.!