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

This one is pretty straightforward if you know that the derivative of `"arcsec"(x)` is equal to

`d/dx("arcsec"(x)) = 1/(|x| * sqrt(x^2-1))`

In your case, you would have to use the chain rule for `"arcsec"(u)`, with `u = x/3` to get

`d/dx(y) = d/(du)("arcsec"(u)) * d/dx(u)`

`y^' = 1/(|u| * sqrt(u^2 - 1)) * d/dx(x/3)`

`y^' = 1/(|x/3| * sqrt((x/3)^2-1)) * (1/3)`

This can be written as

`y^' = color(red)(cancel(color(black)(3)))/(|x| * sqrt((x^2 - 9)/9)) * 1/color(red)(cancel(color(black)(3)))`

`y^' = 1/(1/3 * |x| * sqrt(x^2-9)) = color(green)(3/(|x| * sqrt(x^2-9)))`

A quick word about why the derivative of `"arcsec"(x)` uses the absolute value of `x`.

The idea is that when you use implicit differentiation to find the derivative of `"arcsec"(x)`, you have

`y = "arcsec"(x)" "`, which is equivalent to saying that

`x = secy`

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

`d/dx(x) = d/(dy)(secy) * d/dx(y)`

`1 = secy * tany * (dy)/dx`

Rearrange to get `(dy)/dx` isolated on one side of the equation

`(dy)/dx = 1/(secy * tany)`

Now, for the range of `"arcsec"(x)`, which is `[0, pi/2) uu (pi/2, pi]`, you get that `secy * tany` is always positive.

That happens because you multiply two positive numbers when you have `y in [0, pi/2)` and two negative numbers when you have `y in (pi/2, pi]`.

This will allow you to use

`color(blue)(a = sqrt(a^2)" "(AA)a in [0, +oo))`

Long story short, you can then go ahead and write

`(dy)/dx = 1/sqrt(sec^2y * tan^2y)`

which if you use `tan^2y = sec^2y - 1` will get you

`(dy)/dx = 1/(|sec^2y| * sqrt(sec^2y-1)) - 1/(|x| * sqrt(x^2-1))`