What is the derivative of #arcsec(x/2)#?

Question

2894 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-29T22:03:22

#d/dx (arcsec(x/2) ) = 2/(absxsqrt(x^2-4))#

Let #y = arcsec (x/2)#, then:

#2secy = x#

Differentiate implicitly:

#2secy tany dy/dx = 1#

#dy/dx = 1/(2secy) 1/tany#

#dy/dx = 1/x 1/tany#

using now the identity:

#tan^2y = sec^2y -1 = x^2/4 -1 =(x^2-4)/4#

we have that for #x in [2,+oo)#, that is for #y in [0,pi/2)#:

#tan y = sqrt(x^2-4)/2#, so:

#dy/dx = 2/(xsqrt(x^2-4))#

while for #x in (-oo,-2]#, that is for #y in (pi/2,pi]#:

#tan y = -sqrt(x^2-4)/2#, so:

#dy/dx = -2/(xsqrt(x^2-4))#

We can then write the derivative for both intervals as.

#dy/dx = 2/(absxsqrt(x^2-4))#