# What is the derivative of x tan^-1 - ln sqrt(1+x^2)?

Feb 5, 2017

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

#### Explanation:

$y = x {\tan}^{-} 1 \left(x\right) - \ln \sqrt{1 + {x}^{2}}$

First rewrite the logarithm using $\ln \left({a}^{b}\right) = b \ln \left(a\right)$:

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

Now when we differentiate, we will use the product rule for $x {\tan}^{-} 1 \left(x\right)$ and the chain rule for $\ln \left(1 + {x}^{2}\right)$.

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

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\tan}^{-} 1 \left(x\right)$