University Calculus: Early Transcendentals (3rd Edition)

$2 ( \theta +1) \tanh^{-1} (\theta +1) -1$
We are given: $y=(\theta^2 + 2\theta) \tanh^{-1} (\theta +1)$ Recall the formula: $\dfrac{d (\tanh^{-1} x)}{dx}=\dfrac{1}{1-x^2}$ We need to use the product rule to get the differentiation: $\dfrac{dy}{d \theta}=\tanh^{-1} (\theta +1) (2 \theta +2)+(\theta^2+2 \theta) \dfrac{1}{1- (\theta+1)^2}=\tanh^{-1} (\theta +1) (2 \theta +2)+(\theta^2+2 \theta) \dfrac{1}{1- (\theta^2 + 2 \theta +1)}= 2 ( \theta +1) \tanh^{-1} (\theta +1) -1$