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?
 A: $$\dfrac{d(6\text{arcsec}\ 2x)}{dx}=6\dfrac{d(6\text{arcsec}\ 2x)}{d(2x)}\cdot\dfrac{d(2x)}{dx}=6\cdot\dfrac2{|2x|\sqrt{(2x)^2-1}}$$
A: First, your second line is incorrect.  You're finding the derivative of $6 \operatorname{arcsec}(2x)$, so saying $6 \cdot \dfrac1{x\sqrt{x^2-1}} \cdot 2x'$ is incorrect.  Just remove that second line completely.
Next, the derivative of $\operatorname{arcsec}x$ is $\dfrac1{|x|\sqrt{x^2-1}}$.  The absolute value is necessary in the denominator.  In your case, this will proceed as follows:
\begin{align*}
  \frac d{dx} 6\operatorname{arcsec}(2x) &= 6 \cdot \frac1{|2x|\sqrt{(2x)^2-1}} \cdot \frac d{dx} (2x)\\[0.3cm]
    &= \frac6{2|x|\sqrt{4x^2-1}} \cdot 2\\[0.3cm]
    &= \frac6{|x|\sqrt{4x^2-1}}
\end{align*}
You were close.  Just remove your second line, and put absolute values around the $2x$ in your denominator.
Side note, if you're entering this into an online homework system then you need to be careful with quotients.  If you enter this:  6/x(sqrt(4x^2-1), then first of all the parentheses are unbalanced, but if you fix those and enter this: 6/x(sqrt(4x^2-1)), then the system will interpret it as $\dfrac6x \cdot \sqrt{4x^2-1}$ (every online homework system I know will interpret like this because that's technically the correct interpretation from order of operations).
A: 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}
