Differentiating arctangent directly
We will show by direct arguments that the inverse tangent function is differentiable at all values in its domain. The argument relies on the continuity of
the square root function.
First recall that if 0 ≤ θ < π/2 then sin(θ) ≤ θ ≤ tan(θ). Suppose then that
0 ≤ x < y. Then letting
θ = arctan(y) − arctan(x)
and recalling that
u
1 + u2
1
cos(arctan(u)) = √
1 + u2
sin(a − b) = sin(a) cos(b) − sin(b) cos(a)
tan(a) − tan(b)
tan(a − b) =
1 + tan(a) tan(b)
sin(arctan(u))
√
=
we see
y−x
p
1 + x2 1 + y 2
1
p
√
2
1 + x 1 + y2
√
≤ arctan(y) − arctan(x) ≤
≤
y−x
1 + xy
1
arctan(y) − arctan(x)
≤
.
y−x
1 + xy
It then follows from the Pinching Principle that the inverse tangent function
is differentiable on (0, ∞) and has a righthand derivative at 0. Since the inverse tangent function is an odd function, this is sufficient to prove that it is
differentiable on the whole real line and that
d
1
arctan(x) =
.
(1)
dx
1 + x2
In light of the identities
arccot(x)
=
arcsin(x)
=
arccos(x) =
arcsec(x) =
arccsc(x) =
π
− arctan(x)
2
x
arctan √
1 − x2
π
− arcsin(x)
2
1
arccos
x
π
− arcsec(x)
2
(1) implies that each inverse trigonometric function is differentiable on the interior of its domain. Consequently, each derivative can then be computed by
implicit differentiation or the chain rule.
1