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

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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.

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

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