Answer
$$
f^{\prime}(x)=\frac{d}{dx} \left[ f(x) \right] = \frac{d}{dx} \left[ \tanh \sqrt{x}\right]=\frac{\operatorname{sech}^{2} \sqrt{x}}{2 \sqrt{x}}
$$
Work Step by Step
$$
f(x)=\tanh \sqrt{x}
$$
Differentiating both sides of this equation we have
$$
\begin{aligned}
\frac{d}{dx} \left[ f(x) \right] &= \frac{d}{dx} \left[ \tanh \sqrt{x}\right] \\
f^{\prime}(x) &=\operatorname{sech}^{2} \sqrt{x} \frac{d}{d x} \sqrt{x} \\
&=\operatorname{sech}^{2} \sqrt{x}\left(\frac{1}{2 \sqrt{x}}\right) \\
&=\frac{\operatorname{sech}^{2} \sqrt{x}}{2 \sqrt{x}}
\end{aligned}
$$
