What is the derivative of #f(x)=sec^-1(x)# ?

Question

Charles Anctil · Accepted Answer

#d/dx[sec^-1x] = 1/(sqrt(x^4 - x^2))# Process: First, we will make the equation a little easier to deal with. Take the secant of both sides: #y = sec^-1 x# #sec y = x# Next, rewrite in terms of #cos#: #1/cos y = x# And solve for #y#: #1 = xcosy# #1/x = cosy# #y = arccos(1/x)# Now this looks much easier to differentiate. We know that
#d/dx[arccos(alpha)] = -1/(sqrt(1-alpha^2))#
so we can use this identity as well as the chain rule: #dy/dx = -1/sqrt(1 - (1/x)^2) * d/dx[1/x]# A bit of simplification: #dy/dx = -1/sqrt(1 - 1/x^2) * (-1/x^2)# A little more simplification: #dy/dx = 1/(x^2sqrt(1 - 1/x^2))# To make the equation a little prettier I will move the #x^2# inside the radical: #dy/dx = 1/(sqrt(x^4(1 - 1/x^2)))# Some final reduction: #dy/dx = 1/(sqrt(x^4 - x^2))# And there's our derivative. When differentiating inverse trig functions, the key is getting them in a form that's easy to deal with. More than anything, they're an exercise in your knowledge of trig identities and algebraic manipulation.

Isaiah Allen · Answer

undefined The derivative of $ f(x) = \sec^{-1}(x) $ is:

$$ f'(x) = \frac{1}{|x|\sqrt{x^2 - 1}} $$