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