## Calculus: Early Transcendentals 8th Edition

$y'=\dfrac{\sqrt{x}}{2(1-x)}+\tanh^{-1}\sqrt{x}$
$y=x\tanh^{-1}\sqrt{x}$ 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: $...=\dfrac{1}{2}x\Big(\dfrac{1}{1-x}\Big)(x^{-1/2})+\tanh^{-1}\sqrt{x}=...$ $...=\dfrac{x^{1/2}}{2(1-x)}+\tanh^{-1}\sqrt{x}=\dfrac{\sqrt{x}}{2(1-x)}+\tanh^{-1}\sqrt{x}$