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

##### 1 Answer

#### Explanation:

This one is pretty straightforward **if** you know that the derivative of

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

In your case, you would have to use the chain rule for

#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

The idea is that when you use implicit differentiation to find the derivative of

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

#x = secy#

Use implicit differentiation to differentiate this with respect to

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

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

Rearrange to get

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

Now, for the **range** of **always** positive.

That happens because you multiply two positive numbers when you have

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

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