Differentiate the function with respect to $x$: $\sec (\tan (\sqrt{x}))$

(N/A) Let $f(x) = \sec (\tan (\sqrt{x}))$.
Using the chain rule,we differentiate the function step by step.
Let $u = \sqrt{x}$,$v = \tan(u) = \tan(\sqrt{x})$,and $w = \sec(v) = \sec(\tan(\sqrt{x}))$.
Then,$\frac{df}{dx} = \frac{dw}{dv} \cdot \frac{dv}{du} \cdot \frac{du}{dx}$.
First,$\frac{dw}{dv} = \frac{d}{dv}(\sec(v)) = \sec(v) \tan(v) = \sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x}))$.
Second,$\frac{dv}{du} = \frac{d}{du}(\tan(u)) = \sec^2(u) = \sec^2(\sqrt{x})$.
Third,$\frac{du}{dx} = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$.
Multiplying these together,we get:
$\frac{df}{dx} = \sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x})) \cdot \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$.
Thus,the derivative is $\frac{\sec^2(\sqrt{x}) \sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x}))}{2\sqrt{x}}$.