Single Variable Calculus, Chapter 7, Review Exercises, Section Review Exercises, Problem 46

Differentiate $\displaystyle y = x \tan h^{-1} \sqrt{x}$


$
\begin{equation}
\begin{aligned}

y' =& \frac{d}{dx} (x \tan h^{-1} \sqrt{x})
\\
\\
y' =& x \frac{d}{dx} (\tan h^{-1} \sqrt{x}) + \tan h^{-1} \sqrt{x} \frac{d}{dx} (x)
\\
\\
y' =& x \cdot \frac{1}{1 - (\sqrt{x})^2} \frac{d}{dx} (\sqrt{x}) + \tan h^{-1} \sqrt{x}
\\
\\
y' =& \frac{x }{1 - x} \frac{d}{dx} (x^{\frac{1}{2}}) + \tan h^{-1} \sqrt{x}
\\
\\
y' =& \frac{x}{1 - x} \cdot \frac{1}{2} x^{\frac{-1}{2}} + \tan h^{-1} \sqrt{x}
\\
\\
y' =& \frac{x}{2x^{\frac{1}{2}} (1 - x)} + \tan h^{-1} \sqrt{x}
\\
\\
y' =& \frac{x^{\frac{1}{2}}}{2(1 - x)} + \tan h^{-1} \sqrt{x}
\\
\\
& \text{or}
\\
\\
y' =& \frac{\sqrt{x}}{2 (1 - x)} + \tan h^{-1} \sqrt{x}
\end{aligned}
\end{equation}
$