Differentiate the functions with respect to x.

Let $f(x)=\sec (\tan \sqrt{x}), u(x)=\sqrt{x}, v(t)=\tan t$, and $w(s)=\sec s$

Then, $($ wovou $)(x)=w[v(u(x))]=w[v(\sqrt{x})]=w(\tan \sqrt{x})=\sec (\tan \sqrt{x})=f(x)$

Thus, f is a composite function of three functions, u, v, and w.

Put $s=v(t)=\tan t$ and $t=u(x)=\sqrt{x}$

Then, $\frac{d w}{d s}=\frac{d}{d s}(\sec s)=\sec s \tan s=\sec (\tan t) \cdot \tan (\tan t) \quad[s=\tan t]$

$=\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \quad[t=\sqrt{x}]$

$\frac{d s}{d t}=\frac{d}{d t}(\tan t)=\sec ^{2} t=\sec ^{2} \sqrt{x}$

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

Hence, by chain rule, we obtain

$\frac{d t}{d x}=\frac{d w}{d s} \cdot \frac{d s}{d t} \cdot \frac{d t}{d x}$

$=\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \times \sec ^{2} \sqrt{x} \times \frac{1}{2 \sqrt{x}}$

$=\frac{1}{2 \sqrt{x}} \sec ^{2} \sqrt{x} \sec (\tan \sqrt{x}) \tan (\tan \sqrt{x})$

$=\frac{\sec ^{2} \sqrt{x} \sec (\tan \sqrt{x}) \tan (\tan \sqrt{x})}{2 \sqrt{x}}$

Alternate method

$\frac{d}{d x}[\sec (\tan \sqrt{x})]=\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \frac{d}{d x}(\tan \sqrt{x})$

$=\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \sec ^{2}(\sqrt{x}) \cdot \frac{d}{d x}(\sqrt{x})$

$=\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \cdot \sec ^{2}(\sqrt{x}) \cdot \frac{1}{2 \sqrt{x}}$

$=\frac{\sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \sec ^{2}(\sqrt{x})}{2 \sqrt{x}}$