Answer
$y'=\dfrac{\sqrt{x}}{2(1-x)}+\tanh^{-1}\sqrt{x}$
Work Step by Step
$y=x\tanh^{-1}\sqrt{x}$
Start the differentiation process by using the product rule:
$y'=x(\tanh^{-1}\sqrt{x})'+(\tanh^{-1}\sqrt{x})(1)=...$
$...=x(\tanh^{-1}\sqrt{x})'+\tanh^{-1}\sqrt{x}=$
Apply the chain rule to evaluate the indicated derivative:
$...=(x)\Big[\dfrac{1}{1-(\sqrt{x})^{2}}\Big](\sqrt{x})'+\tanh^{-1}\sqrt{x}=...$
$...=(x)\Big(\dfrac{1}{1-x}\Big)(x^{1/2})'+\tanh^{-1}\sqrt{x}=...$
Evaluate the remaining derivative and simplify:
$...=\dfrac{1}{2}x\Big(\dfrac{1}{1-x}\Big)(x^{-1/2})+\tanh^{-1}\sqrt{x}=...$
$...=\dfrac{x^{1/2}}{2(1-x)}+\tanh^{-1}\sqrt{x}=\dfrac{\sqrt{x}}{2(1-x)}+\tanh^{-1}\sqrt{x}$
