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Файл:Lectures on differential calculus of functions of one variable. Textbook
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17. Properties of differentiable functions 141
Proof.
Let us represent the function xαas a superposition:
xα= e
α ln x
= (ey) ◦(α ln x).
Since we have already proved the differentiability of the functions eyand
ln x, we can apply the superposition differentiation theorem:
(xα)0=(ey) ◦(α ln x)
0
= (ey)0|
y=α ln x
(α ln x)0=
= ey|
y=α ln x
α
x
= αe
α ln x
1
x
= αxαx−1= αx
α−1
.
2. The power-exponential function f(x)
g(x)
, which is defined for an arbitrary function g(x) and a function f (x) that takes positive values, f(x) > 0,
is differentiable at any point of its domain of definition if the functions f and
g are differentiable at this point, and for its derivative the formula holds:
f(x)
g(x)
0
= g0(x)f(x)
g(x)
ln f (x) + f0(x)g(x)f (x)
g(x)−1
.
Remark.
To remember this formula, it is enough to notice that its first term on the
right-hand side can be obtained by differentiating an exponential function
of the form a
g(x)
(that is, (a
g(x))0
= g0(x)a
g(x)
ln a), after which the base a
is replaced by f(x), and the second term can be obtained by differentiating
a power function of the form f (x)α(that is, (f(x)α)0= f0(x)αf(x)
α−1
), after
which the exponent α is replaced by g(x).
Proof.
Let us represent a power-exponential function as f (x)
g(x)
= e
g(x) lnf (x)
and
use the superposition differentiation theorem:
f(x)
g(x)
0
= (e
g(x) lnf (x))0
= (ey) ◦g(x) ln f (x)=
= (ey)0|
y=g(x) ln f(x)
g(x) ln f (x)
0
.
Since (ey)0= ey,
(ey)0|
y=g(x) ln f(x)
= e
g(x) lnf (x)
= f(x)
g(x)
.
We transform the factorg(x) ln f (x)
0
separately using the formula of the
derivative of product and the formula of the derivative of superposition:
g(x) ln f (x)
0
= g0(x) ln f(x) + g(x)ln f (x)
0
=
= g0(x) ln f(x) + g(x) (lny)0|
f(x)
f0(x) =
= g0(x) ln f(x) +
f0(x)g(x)
f(x)
.
142 M. E. Abramyan. Lectures on differential calculus
Let us multiply the resulting expressions:
f(x)
g(x)
0
= f(x)
g(x)
g0(x) ln f(x) +
f0(x)g(x)
f(x)
=
= g0(x)f(x)
g(x)
ln f (x) +
f0(x)g(x)f (x)
g(x)
f(x)
.
Taking into account that
f(x)
g(x)
f(x)
= f(x)
g(x)−1
, we obtain the formula to be
proved.
Differentiation of inverse function
Theorem on the differentiation
of an inverse function 16B/38:57 (04:55), 17A/00:00 (14:50)
Theorem (on the differentiation of inverse function).
Let the function f be continuous and strictly monotone on the segment
[a, b], be differentiable at the point x0∈ (a, b) and f0(x0) 6= 0. By virtue
of continuity and strict monotonicity, the function f has the inverse function f−1(y) defined and continuous on the segment [c,d] = f([a, b]), and
the interval (c, d) contains the point y0= f(x0). Then the function f−1is
differentiable at the point y0and
f−1(y0)
0
=
1
f0(x0)
. (6)
Proof8.
Let us prove the validity of formula (6) for the derivative of the function f−1at the point y0, which immediately implies the differentiability of
the function f−1at a given point.
Let us write the definition of the derivative of f−1at the point y0:
f−1(y0)
0
= lim
y→y
0
f−1(y) −f−1(y0)
y − y
0
=
1
lim
y→y
0
y−y
0
f−1(y)−f−1(y0)
. (7)
The fraction in the denominator makes sense in the neighborhood of the
point y0due to the strict monotonicity of the function f−1: if y 6= y0, then
f−1(y) 6= f−1(y0), therefore, the denominator of the fraction does not vanish.
Using the fact that y = ff−1(y)and f(x0) = y0, we represent the
fraction in the denominator as a superposition:
1
lim
y→y
0
y−y
0
f−1(y)−f−1(y0)
=
1
lim
y→y
0
f(x)−f(x0)
x−x
0
◦ f−1(y)
.
8
This version of the proof is slightly different from the version of the video lectures.
17. Properties of differentiable functions 143
Since f−1(y) is continuous at y0, we obtain that
lim
y→y
0
f−1(y) = f−1(y0) = x0.
The limit of the external function
f(x)−f(x0)
x−x
0
, as x → x0, exists, since the
function f (x) has a derivative at the point x0. In addition, due to the strict
monotonicity of the function f−1, if y 6= y0, then f−1(y) 6= f−1(y0). Thus,
all the conditions of the limit superposition theorem are satisfied, the limit
of superposition exists and is equal to the limit of the external function:
1
lim
y→y
0
f(x)−f(x0)
x−x
0
◦ f−1(y)
=
1
lim
x→x
0
f(x)−f(x0)
x−x
0
=
1
f0(x0)
.
Therefore, the limit (7), from which we started the transformation, also
exists and is equal to
1
f0(x0)
. Formula (6) and, thus, the differentiability of the
inverse function f−1(y) at the point y0are proved.
Remarks.
1. Formula (6) of the derivative of the inverse function at the point y0can
also be written as follows, without using the notation x0:
f−1(y0)
0
=
1
f0(x)|
x=f−1(y0)
. (8)
One can also specify the expression f−1(y0) as an argument to the function f0, but in this case it is desirable to clarify that differentiation is carried
out with respect to the variable x:
f−1(y0)
0
=
1
f
0
x
f−1(y0)
.
2. Under the assumption that the differentiability of the inverse function
has already been proved, formula (6) can be easily obtained from the superposition differentiation theorem. Consider the identity y = ff−1(y)and
find the derivative of both its parts at the point y0. On the left-hand side
we obtain 1, and on the right-hand side we differentiate a superposition as
follows:
1 =ff−1(y0)
0
= f0(x)|
x=f−1(y0)
f−1(y0)
0
.
If we divide the left-hand and right-hand sides of the resulting equality by
f0(x)|
x=f−1(y0)
, we obtain formula (8), which is one of versions of formula (6).
144 M. E. Abramyan. Lectures on differential calculus
Corollaries: derivatives
of inverse trigonometric functions 17A/14:50 (13:12)
The inverse function differentiation theorem simplifies finding derivatives
for functions that are inverse to those elementary functions for which derivative formulas are already known. As an example, we find formulas for derivatives of inverse trigonometric functions.
1. The function arcsin y acts from [−1, 1] to−
π
2
,
π
2
and is monotonically increasing and continuous. According to the inverse function differentiation theorem, the derivatives for the function arcsin y at the point y and
for the function sinx at the point x = arcsin y are related by the equality
(arcsin y)0=
1
(sin x)
0
provided that (sin x)06= 0.
Since (sin x)0= cos x, and the equality cos x = 0 holds only at x = ±
π
2
on the segment−
π
2
,
π
2
, we obtain that the arcsine derivative exists at all
points y such that arcsin y 6= ±
π
2
, that is, at all points of the interval (−1, 1).
Assuming y ∈ (−1, 1) we get the formula for the derivative of arcsin y:
(arcsin y)0=
1
(sin x)0|
x=arcsin y
=
1
cos(arcsin y)
.
For x ∈−
π
2
,
π
2
, the value of cos x is greater than 0, so from the
Pythagorean trigonometric identity we obtain: cos x =p1 −sin2x. In addition, since the functions sine and arcsine are mutually inverse, the identity
sin(arcsin y) = y holds for all y ∈ (−1, 1). Therefore,
1
cos(arcsin y)
=
1
p
1 −sin2(arcsin y)
=
1
p
1 −y
2
.
So, the formula for the arcsine derivative is as follows:
(arcsin y)0=
1
p
1 −y
2
.
This formula makes sense for all y ∈ (−1, 1). The derivative (arcsin y)
0
approaches +∞ as y → ±1.
2. The function arccos y acts from [−1, 1] to [0, π] and is monotonically
decreasing and continuous. Its derivative exists for y ∈ (−1, 1) and is calculated by the formula
(arccos y)0= −
1
p
1 −y
2
.
This formula can be proved in the same way as the formula for the arcsine
derivative, given that (cos x)0= −sin x.
17. Properties of differentiable functions 145
3. The function arctany is defined for all y ∈ R, takes values on the
interval−
π
2
,
π
2
, and is monotonically increasing and continuous. Since the
function tan x has a derivative that is not equal to 0 on the interval−
π
2
,
π
2
,
(tan x)0=
1
cos2x
6= 0, x ∈−
π
2
,
π
2
, the function arctan y is differentiable at
any point y ∈ R:
(arctan y)0=
1
(tan x)0|
x=arctan y
=
1
1
cos2(arctan y)
=
= cos2(arctan y) =
1
1 + tan2(arctan y)
=
1
1 + y
2
.
In deriving this formula, we used the relation
1
cos2x
= 1 + tan2x, which
follows from the Pythagorean trigonometric identity.
So, the formula for the arctangent derivative is as follows:
(arctan y)0=
1
1 + y
2
.
This formula makes sense for all y ∈ R. The derivative (arctan y)0ap-
proaches 0 as y → ±∞.
Remark.
It is interesting to note that the derivative of arctangent, like the derivative
of the logarithm, is a rational function, although the original functions are
not rational.
18. Hyperbolic and inverse hyperbolic
functions
Hyperbolic functions
and their properties 17A/28:02 (11:04)
Definition.
The functions hyperbolic sine (notation sinh x) and hyperbolic cosine (no-
tation cosh x) are defined as follows:
sinh x
def
=
ex− e
−x
2
, cosh x
def
=
ex+ e
−x
2
.
The hyperbolic tangent function (notation tanh x) is the ratio of the hy-
perbolic sine to the hyperbolic cosine:
tanh x
def
=
sinh x
cosh x
=
ex− e
−x
ex+ e
−x
.
Although the definitions of the hyperbolic sine and cosine do not resemble
the definitions of the “ordinary” trigonometric functions sin x and cos x, many
properties of hyperbolic functions are similar to the properties of trigometric
functions.
Consider the basic properties of hyperbolic functions (Fig. 8).
Fig. 8. Graphs of hyperbolic functions
The hyperbolic cosine function is an even function, it is 1 at the point 0:
cosh 0 =
e0+e
0
2
= 1. The hyperbolic cosine approaches +∞ as x → ±∞ and
it grows as an exponential function, that is, faster than any power function.
18. Hyperbolic and inverse hyperbolic functions 147
The hyperbolic sine function is an odd function, it is 0 at the point 0:
sinh 0 =
e0−e
0
2
= 0. The hyperbolic sine approaches ±∞ as x → ±∞ and it
also grows as an exponential function.
It should be noted that the difference cosh x − sinh x is e−x, therefore it
is always positive and cosh x − sinh x → 0 as x → +∞.
The hyperbolic tangent function is an odd function, it is 0 at the point 0:
tanh 0 =
sinh 0
cosh 0
= 0. The hyperbolic tangent approaches ±1 as x → ±∞. For
example, let us prove this for x → +∞:
lim
x→+∞
tanh x = lim
x→+∞
ex− e
−x
ex+ e
−x
= lim
x→+∞
1 −e
−2x
1 + e
−2x
= 1.
For hyperbolic functions, there exists an analogue of the Pythagorean
trigonometric identity:
cosh2x − sinh2x = 1. (1)
This relation can be proved directly using the definitions of the functions
sinh x and cosh x.
Let us find the derivatives of hyperbolic functions:
(sinh x)0=
(ex− e−x)
0
2
=
ex+ e
−x
2
= cosh x,
(cosh x)0=
(ex+ e−x)
0
2
=
ex− e
−x
2
= sinh x.
To find the derivative of the hyperbolic tangent, we use the formulas al-
ready found for the derivatives sinh x and cosh x, as well as the relation (1):
(tanh x)0=
sinh x
cosh x
0
=
(sinh x)0cosh x − sinh x(coshx)
0
cosh2x
=
=
cosh2x − sinh2x
cosh2x
=
1
cosh2x
.
Thus, the formulas of the derivatives for hyperbolic functions are very
similar to the formulas of the derivatives of trigonometric functions.
Inverse hyperbolic functions
and their properties 17B/00:00 (16:45)
Graphs of hyperbolic functions (see Fig. 8 in the previous section) allow us
to assume that the functions sinh x and tanh x are one-to-one, and, therefore,
there exist inverse functions for them. Let us derive a formula for the inverse
function to the hyperbolic sine. To do this, we express the variable x through
the variable y in the equation sinh x = y:
148 M. E. Abramyan. Lectures on differential calculus
sinh x = y,
ex− e
−x
2
= y, ex− e−x= 2y.
Move the 2y term to the left and multiply the resulting equality by ex:
ex− 2y − e−x= 0, e2x− 2yex− 1 = 0.
If we make the change of variables t = ex, then the last equation takes the
form:
t2− 2yt − 1 = 0.
Let us find the roots of the obtained quadratic equation:
t
1,2
= y ±py2+ 1. Since t = ex, we are only interested in the posi-
tive root:
y +py2+ 1 = ex, x = lny +py2+ 1.
We have obtained the formula for the inverse function to the hyperbolic
sine, which makes sense for all y ∈ R. The function that is inverse to the
hyperbolic sine is called the areasine and is denoted by arsinh y. Thus, the
function arsinh y acts from R to R and is expressed by the following formula:
arsinh y = lny +py2+ 1. (2)
Similarly, solving the equation tanh x = y relative to x, we can obtain the
formula for the inverse function to the hyperbolic tangent:
x =
1
2
ln
1 + y
1 −y
.
The function inverse to the hyperbolic tangent is called the areatangent
and denoted by artanh y. Since the function tanh x acts from R to (−1, 1),
we obtain that the function artanh y acts from (−1, 1) to R and is expressed
by the following formula:
artanh y =
1
2
ln
1 + y
1 −y
. (3)
The function cosh x does not have an inverse function on the entire numerical axis R, since the equality cosh x = cosh(−x) holds for all x 6= 0. However,
the restriction of the function cosh x to the half axis [0, +∞) has the inverse
function. This function is called the areacosine, denoted by arcosh y, and
acts from [1, +∞) to [0, +∞) by the formula, which can be obtained in the
same way as the formula for the areasine:
arcosh y = lny +py2− 1. (4)
18. Hyperbolic and inverse hyperbolic functions 149
Since the derivatives of the functions sinh x and tanh x do not vanish anywhere, we obtain, by virtue of the theorem on the differentiation of the inverse
function, that the functions arsinhy and artanhy are also differentiable at
all points of their domain of definition. The derivative of the function cosh x
vanishes at x = 0; therefore, the function arcosh y is non-differentiable at
the point y = cosh 0 = 1, however, it is differentiable at other points in its
domain of definition, that is, y ∈ (1, +∞).
Derivatives of inverse hyperbolic functions can be found by differentiating
formulas (2), (3), (4), which expess these functions by means of other elementary functions, and using the theorem on differentiation of superposition.
As an example, let us find the derivative arsinh y in this way:
(arsinh y)0= (lny +py2+ 1)0= (ln t ◦y +py2+ 1)0=
=
1
t
y+√y2+1
1 +
2y
2py2+ 1
!
=
=
1
y +py2+ 1
p
y2+ 1 + y
p
y2+ 1
=
1
p
y2+ 1
.
Thus,
(arsinh y)0=
1
p
1 + y
2
. (5)
This formula makes sense for all y ∈ R and differs from the formula for the
derivative of the function arcsin only in the minus sign in the denominator.
Formula (5) can also be obtained in a simpler way, if we use the theorem
on the differentiation of the inverse function, the formula for the derivative
sinh x, and relation (1):
(arsinh y)0=
1
(sinh x)
0
x=arsinh y
=
1
cosh x
x=arsinh y
=
=
1
cosh(arsinh y)
=
1
q
1 + sinh2(arsinh y)
=
1
p
1 + y
2
.
When finding the derivative for the function arcosh y in the same way,
we must take into account that this function takes non-negative values, and
therefore, in the formula sinh x = ±pcosh2x − 1, which follows from rela-
tion (1), we should take the plus sign:
(arcosh y)0=
1
(cosh x)
0
x=arcosh y
=
1
sinh x
x=arcosh y
=
150 M. E. Abramyan. Lectures on differential calculus
=
1
sinh(arcosh y)
=
1
q
cosh2(arcosh y) − 1
=
1
p
y2− 1
.
Thus,
(arcosh y)0=
1
p
y2− 1
.
This formula makes sense for all y ∈ (1, +∞). The derivative of the
function arcosh y approaches infinity as y → 1.
Let us also give a formula for the derivative of the function artanh y, which
can be obtained in a similar way using the derivative of the function tanh x
and relation (1):
(artanh y)0=
1
1 −y
2
.
This formula makes sense for all y from the scope of (−1, 1) of the function
artanh y. It differs from the formula for the derivative of the function arctan
only in the minus sign in the denominator.
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