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Solution #1

${\tanh }^{-1}\left(\sqrt{x}\right)$$\stackrel{\text{differentiate w.r.t. x}}{\to }$$\frac{1}{2\sqrt{x}\left(1-x\right)}$$$

Solution #2

$\frac{\ⅆ}{\ⅆ x} {\tanh }^{-1}\left(\sqrt{x}\right)$ = $\frac{1}{2\sqrt{x}\left(1-x\right)}$$$

Solution #3 (stepwise)

$$$\frac{\ⅆ}{\ⅆ x} {\tanh }^{-1}\left(\sqrt{x}\right)$$\stackrel{\text{show solution steps}}{\to }$$\begin{array}{lll}& & \text{Differentiation Steps}\\ & & \frac{\ⅆ}{\ⅆx}\mathrm{arctanh}\left(\sqrt{x}\right)\\ \text{▫}& & \text{1. Apply the}\mathbf{\text{chain}}\text{rule to the term}\mathrm{arctanh}\left(\sqrt{x}\right)\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{chain}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}f\left(g\left(x\right)\right)=\mathrm{f\text{'}}\left(g\left(x\right)\right)\frac{\ⅆ}{\ⅆx}g\left(x\right)\\ & \text{◦}& \text{Outside function}\\ & & f\left(v\right)=\mathrm{arctanh}\left(v\right)\\ & \text{◦}& \text{Inside function}\\ & & g\left(x\right)=\sqrt{x}\\ & \text{◦}& \text{Derivative of outside function}\\ & & \frac{\ⅆ}{\ⅆv}f\left(v\right)=\frac{1}{-v^2+1}\\ & \text{◦}& \text{Apply composition}\\ & & \mathrm{f\text{'}}\left(g\left(x\right)\right)=\frac{1}{-x+1}\\ & \text{◦}& \text{Derivative of inside function}\\ & & \frac{\ⅆ}{\ⅆx}g\left(x\right)=\frac{1}{2\sqrt{x}}\\ & \text{◦}& \text{Put it all together}\\ & & \frac{\ⅆ}{\ⅆx}f\left(g\left(x\right)\right)\frac{\ⅆ}{\ⅆx}g\left(x\right)=\frac{1}{-x+1}\cdot \left(\frac{1}{2\sqrt{x}}\right)\\ & & \text{This gives:}\\ & & \frac{1}{2\left(-x+1\right)\sqrt{x}}\end{array}$$$

Figure 2.10.2(a)   Context Panel access to the Differentiation Rules