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

ivette perez

Posted 10 years ago

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?

POSTED BY: ivette perez

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Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 10 years ago

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": ![enter image description here][1] 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.

POSTED BY: Henrik Schachner

Steve Roth

Steve Roth, Home

Posted 1 year ago

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)

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)

POSTED BY: Steve Roth

Michael Rogers

Michael Rogers, Emory University

Posted 1 year ago

POSTED BY: Michael Rogers

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