Why is the derivative of arcsec(x) different on Wolfram?

Questions
1

The derivative of arcsec(x) is 1/(x(rad(x^2-1)) but when I searched for the derivative on Wolfram, the answer I receive is 1/((x^2)(rad(1-(1/x^2))). Why is this? Is there a restriction not stated?

2

Hello Ivette,

you can try the following:

(* Mathematica: *)
deriv = D[ArcSec[x], x];  
(* your definition: *)
someDef = 1/(x (Sqrt[x^2 - 1]));  
(* are they equivalent? *)              
eq = FullSimplify[deriv == someDef]

which gives for “eq”:

deriv.png

From this one can see that both expressions are the same only if x > 0 or - more general - if x is an element of the positive half-plane. “Proof”:

Refine[eq, Re[x] > 0]   (* output: True *)

Henrik

EDIT:

Since the ArcSec has the form ((antisymmetric function) + Pi/2), its derivative must be symmetric. This is the case only in the result of Mathematica.

3

Well, it looks to me as if the statement “expressions are the same only if x > 0” should read “Abs[x] > 1” – which is the domain of ArcSec.

Also, the result for “eq” does not seem helpful because the numerator will always be zero.

I have been trying to do something similar – getting Mathematica to acknowledge that the formula in my (new) textbook matches.

So:

ClearAll[x, h]

df = ArcSec'[x]

dflim = Limit[(ArcSec[x + h] - ArcSec[x])/h, h -> 0]

dftbl = 1/(Abs[x] (x^2 - 1)^(1/2))

df == dflim (yields true)

dftbl == dflim (yields 1/(Sqrt[-1 + x^2]*Abs[x]) == 1/(Sqrt[1 - 1/x^2]*x^2))

simpdftbl = FullSimplify[dftbl, {x > 1 || x < -1}] (yields 1/Sqrt[x^2*(-1 + x^2)])

simpdflim = FullSimplify[dflim, {x > 1 || x < -1}] (yields 1/Sqrt[x^2*(-1 + x^2)])

simpdftbl == simpdflim (yields true)
4

It depends on the definition of ArcSec[x]. The Wolfram definition (for real x, t): t == ArcSec[x] iff Sec[t] == x and t is between 0 and Pi. With that definition, the derivative of ArcSec[x] is positive over its entire real domain. In this range for t, we have Cos[t] == 1/x and Sin[t] == + Sqrt[1-Cos[t]^2] == Sqrt[1 - 1/x^2]. From implicit differentiation, we get \sec t \tan t {dt \over dx} = 1. Solving for dt/dx and substituting the formulas for sine and cosine in terms of x, we have

{dt \over dx} = {\cos^2 t \over \sin t} = {1 \over x^2 \sqrt{\displaystyle 1 - {1\over x^2}}}\,.

To invert the secant function, one picks a domain over which the function is one-to-one. There are other definitions of arc secant (Stewart’s calculus book, for instance) that arise from choosing different domains. The Wolfram definition is standard. The Wolfram formula for the derivative nicely avoids the absolute value that is found in textbooks and websites with the standard calculus definitions and formulas. The Wolfram formula is valid for complex values of x.