# What is the derivative of  x ln(arctan x)?

Mar 17, 2018

The derivative is $= \ln \left(\arctan \left(x\right)\right) + \frac{x}{\left(1 + {x}^{2}\right)} \cdot \frac{1}{\arctan \left(x\right)}$

#### Explanation:

Let $u = \arctan x$

Then,

$\tan u = x$

${\sec}^{2} u = 1 + {x}^{2}$

Differentiating wrt $x$

$\left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \cdot {\sec}^{2} u = 1$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{\sec} ^ 2 u = \frac{1}{1 + {x}^{2}}$

Therefore,

$y = x \ln \left(\arctan x\right)$

$y = x \ln \left(u\right)$

Differentiating wtt $x$ by pplying the product rule

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 \cdot \ln u + x \cdot \frac{\mathrm{du}}{\mathrm{dx}} \cdot \frac{1}{u}$

$= \ln \left(\arctan \left(x\right)\right) + \frac{x}{\left(1 + {x}^{2}\right)} \cdot \frac{1}{\arctan \left(x\right)}$

Mar 17, 2018

$\frac{d}{\mathrm{dx}} x \ln \arctan x = \ln \arctan x + \frac{x}{\left(1 + {x}^{2}\right) \arctan x}$

#### Explanation:

To find $\frac{d}{\mathrm{dx}} x \ln \arctan x$, we use the product rule

$\frac{d}{\mathrm{dx}} u v = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dx}} \ln \arctan x = \frac{1}{\left(1 + {x}^{2}\right) \arctan x}$

So

$\frac{d}{\mathrm{dx}} x \ln \arctan x = \ln \arctan x + \frac{x}{\left(1 + {x}^{2}\right) \arctan x}$