The derivative of the inverse function
Suppose that f is one-to-one and differentiable on an interval (a, b). Suppose
that its inverse function f −1 is also differentiable (this assumption may be
dropped, in that one may prove directly that the inverse function of a differentiable function is automatically differentiable, see *). We differentiate the
equation
x = f (f −1 (x)),
using the Chain rule. The result is
1 = f 0 (f −1 (x)) (f −1 )0 (x).
Hence,
(f −1 )0 (x) =
1
f 0 (f −1 (x))
.
* This makes it clear why an extra condition on f , namely that f 0 (f −1 (x)) 6= 0
is needed to guarantee the differentiability of f −1 at x.
Sine and its inverse function Arcsin
Below are the graphs of sin x on the range [−π/2, π/2], the graph of its inverse
function arcsin x on the interval [−1, 1] and the line y = x.
1.5
1
0.5
-1.5
-1
-0.5
0.5
1
1.5
-0.5
-1
-1.5
Figure 1: Graphs of sin x, of arcsin x and the line y = x
1
We need to calculate the derivative of the function arcsin. By the general
formula derived in the previous section,
,Ã
!¯
¯
d
d
¯
(arcsin x) = 1
sin x ¯¯
dx
dx
arcsin x
= 1 /(cos x)|arcsin x
1
=
.
cos[arcsin x]
In general, cos2 y = 1 − sin2 y and so
cos2 [arcsin x] = 1 − sin2 [arcsin x] = 1 − x2 .
√
Thus, cos[arcsin x] = ± 1 − x2 . We need to decide which sign is correct.
Since arcsin x ∈ [−π/2, π/2] and cosine is non negative in this range, we need
to take the plus sign. Hence,
d
1
(arcsin x) =
,
dx
1 − x2
−1 < x < 1.
Tan and its inverse function Arctan
The function tan x has period π. In fact, sin(x+π) = − sin x and cos(x+π) =
− cos x. Furthermore, neither tan(π/2) nor tan(−π/2) are defined, because
cosine is zero in these cases. On the domain (−π/2, π/2), the function tan x
is one-to-one. One way to see this is to note that the derivative of tan x
is 1/ cos2 x, and this is positive on (−π/2, π/2), whence tan x is strictly increasing on this interval. Furthermore, tan x assumes all real values on the
interval (−π/2, π/2). Restricting the domain of tan x to this interval and
taking the inverse function gives rise to the function arctan. Thus arctan has
domain (−∞, ∞) and range (−π/2, π/2). Below are the graphs of tan and
arctan. We need to calculate the derivative of the function arctan. Again we
use the general formula derived earlier,
d
(arctan x) = 1
dx
Now,
tan2 y =
,Ã
d
tan x
dx
!¯
¯
¯
¯
¯
= cos2 [arctan x].
arctan x
1 − cos2 y
1
sin2 y
=
=
−1
cos2 y
cos2 y
cos2 y
2
10
7.5
5
2.5
-10
-5
5
10
-2.5
-5
-7.5
-10
Figure 2: Graphs of tan x, of arctan x and the line y = x
Thus, cos2 y = 1/(1 + tan2 y) and so cos2 [arctan x] = 1/[1 + tan2 (arctan x)] =
1/(1 + x2 ). This gives,
d
1
(arctan x) =
.
dx
1 + x2
Mathematics MA1051 - Exercise Sheet B2
1* Suppose that f is a continuous function on an interval [x, y] and that f (x) <
f (y). Suppose that x < z < y, that f (z) < f (x) or that f (z) > f (y). Use
the Intermediate Value Theorem to show that f is not one-to-one on [x, y].
Deduce from this that if f is a continuous function on the interval [a, b] and
is one-to-one on this interval, then f is either strictly increasing on [a, b] or
is strictly decreasing on [a, b].
2 Show that if L and M are lines in the plane that are reflections of each other
in the line y = x, then the product of the slopes of L and M is 1.
3 Sketch the graph of the function
½
f (x) =
(x − 1)/2,
4 − x,
3
1 ≤ x < 2,
2 ≤ x ≤ 3.
Is f one-to-one on the interval [1, 3]? If so, find its inverse function and draw
its graph.
4 Show that each function is one-to-one on the specified domain. In each case
find the inverse function explicitly.
x
on (−∞, ∞)
(a)
f1 (x) = √ 2
x +1
(b)
f2 (x) = (1 − 2x)3 on (−∞, ∞)
√
(c)
f3 (x) = 1 − 2x on (−∞, 1/2)
5 Set arccos to be the inverse function of cosine on [0, π]. Sketch the graphs of
cos x on [0, π] and of arccos x on [−1, 1]. Show that
π
arccos x = − arcsin x,
−1 ≤ x ≤ 1.
2
Find the derivative of arccos.
R
6 Let ln x = 1x dt/t, for x > 0, be the natural logarithm function and let us
think of exp x as its inverse function. What are the domain and range of
the functions ln and exp? Draw their graphs. Show that the exponential
d
(exp x) = exp x, using the formula for the derivative of
function satisfies dx
d
the inverse function and dx
(ln x) = 1/x.
7 Prove that
µ
¶
x−1
π
= arctan x − ,
x > −1.
x+1
4
What form does this identity take on (−∞, −1), if any?
arctan
8 Find, for a > 0,
Z
(i)
Find
√
Z
dt
,
a2 − t2
(ii)
a2
dt
.
+ t2
Z
dt
(Hint: complete the square)
− 2t + 3
R
Compare this problem to dt/(t2 − 4t + 3).
(More difficult.) Find
t2
Z
Z
dt
.
1 + t4
√
√
Note that 1 + t4 can be factored as (t2 + 2 t + 1)(t2 − 2 t + 1).
(i)
dt
,
1 + t3
(ii)
4