Let $f(x) = 6\operatorname{arcsec}(2x)$. Find $f'(x)$.

Question

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?

Answer 1

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).

Answer 2

$$\dfrac{d(6\text{arcsec}\ 2x)}{dx}=6\dfrac{d(6\text{arcsec}\ 2x)}{d(2x)}\cdot\dfrac{d(2x)}{dx}=6\cdot\dfrac2{|2x|\sqrt{(2x)^2-1}}$$

Answer 3

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}