Differentiate the following function with respect to $x$
$tan h^{- 1} (3 x + 1)$

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Solution

$\Rightarrow t a n h^{- 1} x = \frac{1}{2} l n \frac{1 + x}{1 - x}$

$\Rightarrow t a n h^{- 1} (3 x + 1) = \frac{1}{2} l n \frac{2 + 3 x}{- 3 x}$

$\Rightarrow \frac{d}{d x} t a n h^{- 1} (3 x + 1) = \frac{1}{2} (\frac{- 3 x}{2 + 3 x}) (\frac{2}{3 x^{2}})$

$\Rightarrow \frac{d}{d x} t a n h^{- 1} (3 x + 1) = \frac{- 1}{x (2 + 3 x)}$

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