# How do I find the derivative of  f(x) = x tan^-1 - ln sqrt(1+x^2)?

Apr 25, 2018

${\tan}^{-} 1 x$.

#### Explanation:

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

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

Using the usual rules of diffn., we get,

$f ' \left(x\right) = x \cdot \frac{d}{\mathrm{dx}} \left\{{\tan}^{-} 1 x\right\} + {\tan}^{-} 1 x \cdot \frac{d}{\mathrm{dx}} \left\{x\right\}$

$- \frac{1}{2} \cdot \frac{d}{\mathrm{dx}} \left\{\ln \left(1 + {x}^{2}\right)\right\}$,

$= x \cdot \frac{1}{1 + {x}^{2}} + {\tan}^{-} 1 x - \frac{1}{2} \cdot \frac{1}{1 + {x}^{2}} \cdot \frac{d}{\mathrm{dx}} \left\{1 + {x}^{2}\right\}$,

$= \frac{x}{1 + {x}^{2}} + {\tan}^{-} 1 x - \frac{1}{\cancel{2}} \cdot \frac{1}{1 + {x}^{2}} \cdot \cancel{2} x$,

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

$\Rightarrow f ' \left(x\right) = {\tan}^{-} 1 x$.

Spread the Joy of Maths!