Derivative of Arctan(x)
Let’s use our formula for the derivative of an inverse function to find the derivative of the inverse of the tangent function: y = tan−1 x = arctan x.
We simplify the equation by taking the tangent of both sides:
y
=
tan−1 x
tan y
=
tan(tan−1 x)
tan y
= x
To get an idea what to expect, we start by graphing the function tan x. Can
you reflect this graph across the line y = x to get an idea of what the graph of
y = tan−1 x looks like?
need graph.
Recall that:
d
tan y
dy
=
=
=
=
=
d sin y
dy cos y
cos y · cos y) − sin y(− sin y)
cos2 y
2
cos y + sin2 y
cos2 y
1
cos2 y
1
sec2 y
(quotient rule)
We can now take the derivative of both sides of the equation to get:
tan y
d
[tan(y)]
dx
d
dy
[tan(y)]
dy
dx
1
dy
cos2 (y) dx
dy
dx
=
=
x
dx
= 1
dx
=
1
=
1
=
cos2 (y)
Or equivalently, y 0 = cos2 y. Unfortunately, we want the derivative as a
function of x, not of y. We must now plug in the original formula for y, which
1
y
(1+x2)1/2
1
x
Figure 1: Triangle with angles and lengths corresponding to those in the example
illustrating differentiation using the inverse function arctan
was y = tan−1 x to get y 0 = cos2 (arctan(x)). This is a correct answer but it can
be simplified tremendously. We’ll use some geometry to simplify it.
In this triangle, tan(y) = x so
arctan(x) = y
The Pythagorian theorem tells us the length of the hypotenuse:
p
h = 1 + x2
and we can now compute
cos(y) = √
1
.
1 + x2
From this, we get
cos2 (y) =
so
√
1
1 + x2
2
=
1
1 + x2
1
dy
=
.
dx
1 + x2
In other words,
1
d
arctan(x) =
.
dx
1 + x2
2