1
151 WebCalc Fall 2002-copyright Joe Kahlig
For those of you who are interested in the different definitions of the derivative of sec−1 x.
For both of these definitions we start with y = sec−1 x. This is the same as saying sec y = x. Now
take a derivative of both side and solve for y0 gives:
1
y0 =
sec y tan y
√
p
Also the trig identity tan2 y = sec2 y − 1 gives tan y = ± sec2 y − 1 = ± x2 − 1. Don’t forget that
sec y = x
Definition 1 (WebCalc):
y = sec−1 x with the domain of |x| ≥ 1 and range [0, π2 ) ∪ ( π2 , π]
For y ∈ [0, π2 ), i.e. 0 ≤ y < π2 , x = sec y ≥ 1. Having x ≥ 1 means that |x| = x since the absolute
√
value of a positive number is itself. On this interval, We also have that tan y > 0. Thus tan y = x2 − 1
Now substituting this into the derivative from above we get:
1
1
1
√
y0 =
= √
=
2
sec y tan y
x x −1
|x| x2 − 1
For y ∈ ( π2 , π], i.e. π2 < y ≤ π, x = sec y ≤ 1. Having x ≤ 1 means that |x| = −x since the absolute
value of a negative
√ number is the opposite of itself. On this interval, we also have that tan y < 0.
Thus tan y = − x2 − 1
Now substituting this into the derivative from above we get:
1
1
1
1
√
√
√
y0 =
=
=
=
2
2
sec y tan y
x(− x − 1)
−x x − 1
|x| x2 − 1
Definition 2 (Stewart):
y = sec−1 x with the domain of |x| ≥ 1 and range [0, π2 ) ∪ [π, 3π
2 ). For both of these intervals tan y > 0
√
so we have tan y = x2 − 1.
substituting this into the derivative from above we get:
1
1
y0 =
= √
sec y tan y
x x2 − 1
Note the two derivatives of the inverse secant function are not equivalent. Nor should we expect them
to be, since the inverse secant function has two different definitions. The graph of y = sec−1 x is listed
below. Definition 1 on the left and Definition 2 on the right.
−1
3π/2
3π/2
π
π
π/2
π/2
1
−1
1