What is the derivative of arcsec(x/2)arcsec(x2)?

1 Answer
Jul 29, 2018

d/dx (arcsec(x/2) ) = 2/(absxsqrt(x^2-4))ddx(arcsec(x2))=2|x|x24

Explanation:

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=sec2y1=x241=x244

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=x242, so:

dy/dx = 2/(xsqrt(x^2-4))dydx=2xx24

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=x242, so:

dy/dx = -2/(xsqrt(x^2-4))dydx=2xx24

We can then write the derivative for both intervals as.

dy/dx = 2/(absxsqrt(x^2-4))dydx=2|x|x24