If f(x) = arcsec(3x), then
sec(f(x)) = sec(arcsec(3x))
sec(f(x)) = 3x
But bear in mind that the right side reduces in this way only if 0 ≤ f(x) ≤ π.
Differentiating both sides using the chain rule gives
sec(f(x)) tan(f(x)) f'(x) = 3
so that
f'(x) = 3 cos(f(x)) cot(f(x))
f'(x) = 3 cos(arcsec(3x)) cot(arcsec(3x))
We *could* stop here, but we can usually simplify these nested trig and inverse trig expressions to end up with an simpler algebraic one. Consider a right triangle with a reference angle measuring θ = f(x) = arcsec(3x). Then sec(θ) = 3x. It follows from the definition of secant, and subsequently the Pythagorean identity, that
• cos(θ) = 1/sec(θ) = 1/(3x)
• sin(θ) = √((3x)² - 1²) = √(9x² - 1)/(3x)
but remember that we assume 0 ≤ θ ≤ π. Over this interval, sin(θ) can be either positive or negative, which we account for by replacing x with |x|, so that
• sin(θ) = √(9x² - 1)/(3|x|)
So, we have
cos(arcsec(3x)) = 1/(3x)
cot(arcsec(3x)) = (1/(3x)) / (√(9x² - 1)/(3|x|)) = |x|/(x √(1 - 9x²))
and so
f'(x) = 3 • 1/(3x) • |x|/(x √(1 - 9x²))
It's easy to show that |x|/x = x/|x|, so we can rewrite this as
f'(x) = 3 • 1/(3x) • x/(|x| √(1 - 9x²))
f'(x) = 1/(|x| √(1 - 9x²))