Question

How do you take the derivative of `sec^-1 (3x^2)`?

Answer 1

`y^' = 2/(x * sqrt(9x^4-1))`

Explanation:

For your function `y = sec^(-1)(3x^2)`, you know that you can write

`sec(y) = 3x^2`

Use implicit differentiation and the fact that

`color(blue)(d/dx(secx) = secx * tanx)`

to write

`d/(dy)(secy) * (dy)/dx = d/dx(3x^2)`

`secy * tany * (dy)/dx = 6x`

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

`(dy)/dx = (6x)/(secy * tany)`

Now, `"arcsec"(x)` has a range of `[0, pi/2) uu (pi/2, pi]`. If you evaluate the product of `secx` and `tanx` over the first interval, `[0, pi/2)`, you will always end up with a positive number because both the value of the tangent, and the value of secant are positive in that interval.

For the second interval, `(pi/2, 0]`, the product will once again be positive, since the tangent and the secant are both negative in that interval.

The `secx * tanx` product will always be positive, which means that you can write

`secx * tanx = sqrt( (secx * tanx)^2)`

Take this back to your derivative calculation

`(dy)/dx = (6x)/sqrt( (secy * tany)^2) = (6x)/(sqrt(sec^2y * tan^2y))`

Another important thing to remember is that

`color(blue)(sec^2x = 1 + tan^2x`

which means that you have

`(dy)/dx = (6x)/sqrt(sec^2y * (sec^2y - 1))`

`(dy)/dx = (6x)/(|sec^2y| * sqrt(sec^2y-1))`

Since `secy = 3x^2`, you get that

`(dy)/dx = (6x)/(underbrace(3x^2)_(color(red)("always positive")) * sqrt((3x^2)^2 - 1))`

Finally, the derivative of `y = sec^(-1)(3x^2)` is equal to

`(dy)/dx = color(green)(2/(x * sqrt(9x^4-1)))`