calculus - How was the differentiation formula derived for a composite trig function?

For example, derivative of this trig function is:

\begin{align}\frac{d}{dx} \left[\sec\left(\frac{x}{12}\right)\right] \
&= \sec\left(\frac{x}{12}\right) * \tan\left(\frac{x}{12}\right) * \frac{d}{dx}\left(\frac{x}{12}\right) \\

&= \sec\left(\frac{x}{12}\right) * \tan\left(\frac{x}{12}\right) * \frac{1}{12}
\end{align}

But I fail to see what differentiation formula is applied here.
It's not chain rule for sure.
If you do it by chain rule and assume it is a power of 1 then it'd look like this:

$$\frac{d}{dx} [sec(\frac{x}{12})]^1 = 1 * (sec(\frac{x}{12}))^0 * \frac{d}{dx} (sec(\frac{x}{12})) = 1 * 1 * sec(\frac{x}{12}) * tan(\frac{x}{12})$$

If chain rule is not applied to this function like this because the function is "composite" which is why it should be done as the first variant, then how was chain rule altered for this function in the first variant?

Answer

Actaully it is the chain rule. You are confusing the chain rule and the power rule. If $h(x)=f(g(x))$, the chain rule states:
$$h'(x)=f'(g(x))g'(x)$$.

The power rule is just a special case of the chain rule. Namely, when $f(x)=x^n$.

To use the chain rule for your example. Let $f(x)=\sec x$ and $g(x)=\frac x{12}$.

$$h(x)=f(g(x))=f\left(\frac x{12}\right)=\sec\frac x{12}$$

So,
$$h'(x)=f'(g(x))g'(x)=\sec\frac x{12}\tan\frac x{12}\cdot\frac 1{12}=\frac1{12}\sec\frac x{12}\tan\frac x{12}.$$