Answer
$-\tanh^{-1} \theta +\dfrac{1}{(1 + \theta)}$
Work Step by Step
Given: $y=(1-\theta) \tanh^{-1} \theta$
Since, $\dfrac{d (\tanh^{-1} x)}{dx}=\dfrac{1}{1-x^2}$
Apply product rule to get the differentiation.
$\dfrac{dy}{d \theta}=\tanh^{-1} \theta (-1)+(1-\theta) \dfrac{1}{1- \theta^2}=-\tanh^{-1} \theta+(1-\theta) [\dfrac{1}{(1- \theta)(1 + \theta)}]$
or, $=-\tanh^{-1} \theta +\dfrac{1}{(1 + \theta)}$
