Let $f(x) = 6\operatorname{arcsec}(2x)$. Find $f'(x)$. Let $f(x) = 6\operatorname{arcsec}(2x)$. Find $f'(x)$.
$$6 \cdot \frac 1{2x\sqrt{(2x)^2-1}} \cdot 2$$
$$\frac{12}{2x\sqrt{4x^2-1}}$$
$$\frac 6 {x\sqrt{4x^2-1}}$$
Why is that wrong?

You see to say that $\displaystyle \sqrt{x^2-1} = \frac 1 2 \sqrt{(2x)^2 - 1}.$ That is incorrect. In fact
$$
\sqrt{x^2-1} = \frac 1 2 \cdot 2\sqrt{x^2-1} = \frac 1 2 \cdot \sqrt{4}\cdot\sqrt{x^2-1} = \frac 1 2 \sqrt{4x^2-4} = \frac 1 2 \sqrt{(2x)^2-4}.
$$
Now let $y = 6\operatorname{arcsec}(2x).$ Then
\begin{align}
\frac y 6 & = \operatorname{arcsec}(2x) \\[10pt]
\sec\frac y 6 & = 2x \\[10pt]
\frac 1 2 \sec\frac y 6 & = x \\[12pt]
\frac 1 2 \left(\sec\frac y 6\right) \cdot \left(\tan\frac y 6 \right) \cdot \frac 1 6 & = \frac{dx}{dy} \\[10pt]
x \cdot \left( \tan \frac y 6 \right) \cdot \frac 1 6 & = \frac{dx}{dy} \\[10pt]
\pm x \cdot \sqrt{\left( \sec\frac y 6 \right)^2 -1} \cdot \frac 1 6 & = \frac{dx}{dy} \\[10pt]
\pm \frac x 6 \cdot \sqrt{(2x)^2 - 1} & = \frac{dx}{dy} \\[10pt]
\frac{dy}{dx} & = \frac{\pm6}{x\sqrt{(2x)^2 - 1}} 
\end{align}
