What is the derivative of $ln (arctan x)$ $ln(arctanx)$ ?

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Answer 1 · 2015-10-18T18:35:51

$= \frac{1}{(1 + x^{2}) ({tan}^{- 1} x)}$ $=1/((1+x^2)(tan^(-1)x))$

$\frac{d}{d x} ln ({tan}^{- 1} x) = \frac{1}{{tan}^{- 1} x} \cdot \frac{d}{d x} {tan}^{- 1} x$ $d/dxln(tan^(-1)x)=1/(tan^(-1)x)*d/dxtan^(-1)x$

$= \frac{1}{{tan}^{- 1} x} \cdot \frac{1}{1 + x^{2}}$ $=1/(tan^(-1)x)*1/(1+x^2)$

What is the derivative of ln(arctanx)ln(arctanx)ln(arctanx)?