Let y = arcsec (x/2)y=arcsec(x2), then:
2secy = x2secy=x
Differentiate implicitly:
2secy tany dy/dx = 12secytanydydx=1
dy/dx = 1/(2secy) 1/tanydydx=12secy1tany
dy/dx = 1/x 1/tanydydx=1x1tany
using now the identity:
tan^2y = sec^2y -1 = x^2/4 -1 =(x^2-4)/4tan2y=sec2y−1=x24−1=x2−44
we have that for x in [2,+oo)x∈[2,+∞), that is for y in [0,pi/2)y∈[0,π2):
tan y = sqrt(x^2-4)/2tany=√x2−42, so:
dy/dx = 2/(xsqrt(x^2-4))dydx=2x√x2−4
while for x in (-oo,-2]x∈(−∞,−2], that is for y in (pi/2,pi]y∈(π2,π]:
tan y = -sqrt(x^2-4)/2tany=−√x2−42, so:
dy/dx = -2/(xsqrt(x^2-4))dydx=−2x√x2−4
We can then write the derivative for both intervals as.
dy/dx = 2/(absxsqrt(x^2-4))dydx=2|x|√x2−4