# How do you find the derivative of y = sec^(-1)(x/3)?

Aug 22, 2015

${y}^{'} = \frac{3}{| x | \cdot \sqrt{{x}^{2} - 9}}$

#### Explanation:

This one is pretty straightforward if you know that the derivative of $\text{arcsec} \left(x\right)$ is equal to

$\frac{d}{\mathrm{dx}} \left(\text{arcsec} \left(x\right)\right) = \frac{1}{| x | \cdot \sqrt{{x}^{2} - 1}}$

In your case, you would have to use the chain rule for $\text{arcsec} \left(u\right)$, with $u = \frac{x}{3}$ to get

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{du}} \left(\text{arcsec} \left(u\right)\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

${y}^{'} = \frac{1}{| u | \cdot \sqrt{{u}^{2} - 1}} \cdot \frac{d}{\mathrm{dx}} \left(\frac{x}{3}\right)$

${y}^{'} = \frac{1}{| \frac{x}{3} | \cdot \sqrt{{\left(\frac{x}{3}\right)}^{2} - 1}} \cdot \left(\frac{1}{3}\right)$

This can be written as

${y}^{'} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}{| x | \cdot \sqrt{\frac{{x}^{2} - 9}{9}}} \cdot \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}$

${y}^{'} = \frac{1}{\frac{1}{3} \cdot | x | \cdot \sqrt{{x}^{2} - 9}} = \textcolor{g r e e n}{\frac{3}{| x | \cdot \sqrt{{x}^{2} - 9}}}$

A quick word about why the derivative of $\text{arcsec} \left(x\right)$ uses the absolute value of $x$.

The idea is that when you use implicit differentiation to find the derivative of $\text{arcsec} \left(x\right)$, you have

$y = \text{arcsec"(x)" }$, which is equivalent to saying that

$x = \sec y$

Use implicit differentiation to differentiate this with respect to $x$ to get

$\frac{d}{\mathrm{dx}} \left(x\right) = \frac{d}{\mathrm{dy}} \left(\sec y\right) \cdot \frac{d}{\mathrm{dx}} \left(y\right)$

$1 = \sec y \cdot \tan y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

Rearrange to get $\frac{\mathrm{dy}}{\mathrm{dx}}$ isolated on one side of the equation

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sec y \cdot \tan y}$

Now, for the range of $\text{arcsec} \left(x\right)$, which is $\left[0 , \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2} , \pi\right]$, you get that $\sec y \cdot \tan y$ is always positive.

That happens because you multiply two positive numbers when you have $y \in \left[0 , \frac{\pi}{2}\right)$ and two negative numbers when you have $y \in \left(\frac{\pi}{2} , \pi\right]$.

This will allow you to use

$\textcolor{b l u e}{a = \sqrt{{a}^{2}} \text{ } \left(\forall\right) a \in \left[0 , + \infty\right)}$

Long story short, you can then go ahead and write

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{{\sec}^{2} y \cdot {\tan}^{2} y}}$

which if you use ${\tan}^{2} y = {\sec}^{2} y - 1$ will get you

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{| {\sec}^{2} y | \cdot \sqrt{{\sec}^{2} y - 1}} - \frac{1}{| x | \cdot \sqrt{{x}^{2} - 1}}$