MAT221: CALCULUS II | Transcendental Functions -Exponential and Logarithmic Functions - Inverse Trigonometric Functions.pdf

MAT221: CALCULUS II
Transcendental Functions
Josophat Makawa
Student - University of Malawi
C. J. P. Makawa (Student-University of Malawi) MAT221: CALCULUS II February 22, 2025 1 / 52

Contents
1 Exponential & Logarithmic Functions
Introduction
Logarithmic Differentiation
Logarithmic Integration
2 Trigonometric Functions
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric Functions
Integrals Involving Inverse Trigonometric Functions

Exponential & Logarithmic Functions
Exponential & Logarithmic Functions

Exponential & Logarithmic Functions Introduction
Introduction
Firstly, it is well known that Exponential and Logarithmic Functions are inverses
of each other. That is, they satisfy the fact that;
f(f−1
(x)) = f−1
(f(x)) = x (1)
That means,
loga(ax
) = aloga(x)
= x
as long as the base a remains constant through out.
In addition, we also know how to change logarithms from one base to the other.
For example, if you want to change a logarithm of x in base a to base b, we use
loga x =
logb x
logb a
(2)

In MAT121, we looked at the differentiation and integration of the basic
exponential and logarithmic functions. Here, we discussed that the derivative of
an exponential function y = ex
is ex
. When the base is e, it is called Natural
Exponential Function.
For any other base a, we found the derivative of the function y = ax
to be ax
ln a.
For logarithms, we discussed that the natural logarithmic function of x is present
as ln x as a short form of loge x.
We also discussed that the derivative of the function y = ln x is 1
x
. And for any
other general base a we first change it to base e using Equation 2 and then
differentiate which yielded the derivative of y = loga x as 1
x ln a
.
On Integration, we found the reverse processes of the above derivatives. And
for most questions we used integration by substitution to solve them.

SUMMARY
d
dx
eu(x)

= eu(x)
×
du
dx
d
dx
au(x)

= au(x)
×
du
dx
d
dx
(ln u(x)) =
1
u(x)
×
du
dx
d
dx
(loga u(x)) =
1
u(x) ln a
×
du
dx







Derivatives
(3)
Z
eu(x)
du = eu
+ c
Z
au(x)
du =
au(x)
ln a
+ c
Z
1
u(x)
du = ln|u(x)|+c







Integrals (4)

Exponential Logarithmic Functions Logarithmic Differentiation
Logarithmic Differentiation
The other thing that we looked at in MAT121 is the Quotient Rule which dealt
with the differentiation of rational functions. However, some rational functions
are so complicated such that using the quotient seems almost impossible.
For example, consider the function below;
y =
x5
(1 − 10x)
√
x2 + 2
The above function can be solved using the quotient rule but the process will be
very complicated to follow. Therefore we will be using the Logarithmic
Differentiation.
In this process of differentiation, we will be introducing natural logarithms both
sides and solve the function implicitly (Check Implicit differentiation too).

Given y = x5
(1−10x)
√
x2+2
, introduce natural logarithms both sides.
ln(y) = ln

x5
(1 − 10x)
√
x2 + 2

We then apply the rules of indices and logarithms where we know that if the
expressions are multiplying, then, their logs are adding and if the expressions are
dividing, then their logs are subtracting, and a power is a product of a log.
ln y = ln x5
− ln

(1 − 10x)
√
x2 + 2

= ln x5
−
h
ln(1 − 10x) + ln(x2
+ 2)
1
2
i
ln y = ln x5
− ln(1 − 10x) −
1
2
ln(x2
+ 2)
After this, we will differentiate implicitly; each term on its own using equation
3 where in simplified terms, for y = ln u, y′
= u′
u
.

Let’s now use the implicit differentiation to differentiate
ln y = ln x5
− ln(1 − 10x) − ln
√
x2 + 2 term by term as follows:
d
dx
(ln y) =
1
y
×
dy
dx
=
y′
y
since
dy
dx
= y′
d
dx
(ln x5
) =
5x4
x
5
=
5
x
d
dx
(ln(1 − 10x)) =
−10
1 − 10x
d
dx

1
2
ln(x2
+ 2)

=
1

2
×

2x
x2 + 2
=
x
x2 + 2
From here, we just need to combine everything and simplify if possible.

Therefore, the derivative of y = x5
(1−10x)
√
x2+2
is
y′
y
=
5
x
−

−10
1 − 10x

−
x
x2 + 2
=
5
x
+
10
1 − 10x
−
x
x2 + 2
If we multiply y both sides,
y′
=

5
x
+
10
1 − 10x
−
x
x2 + 2

× y
But y = x5
(1−10x)
√
x2+2
, if we substitute we get the final derivative of y as,

5
x
+
10
1 − 10x
−
x
x2 + 2

x5
(1 − 10x)
√
x2 + 2


We can see here that it’s slightly simpler to do the differentiation using
logarithms that it could be if we have used the product and quotient rules. In
addition, the answers are also simplified which improves our understand the
process.
The same approach is also used to differentiate expressions in which functions
are raised to other functions. Thus
If y = f(x)g(x)
, then, ln y = ln f(x)g(x)

= g(x) ln f(x) (5)
Then, we apply implicit differentiation and use the product rule on the right
hand side to finish the differentiation process.

Example ( Differentiate y = xx
)
Solution
Firstly we can happily see that both power rule y′
= nxn−1
and the exponential
differentiation y′
= ax
ln a won’t work. So, let’s introduce the logarithms.
ln y = ln xx
= x ln x
Here, we will differentiate ln y implicitly and x ln x using the product rule.
1
y
·
dx
dy
= 1 · ln x +
x ·

1

x

= ln x + 1
Multiplying by y both sides yields
dy
dx
= (ln x + 1)xx

Example (Given that y = (1 − 3x)cos(x)
, find
dy
dx
)
Solution
Introduce logarithms
ln y = ln

(1 − 3x)cos(x)

= cos(x) ln(1 − 3x)
Differentiate
1
y
·
dy
dx
= − sin(x)·ln(1−3x)+cos(x)·
−3
(1 − 3x)
= −

2 sin(x)(1 − 3x) +
3 cos(x)
(1 − 3x)

−

2 sin(x)(1 − 3x) +
3 cos(x)
(1 − 3x)

(1 − 3x)cos(x)


Before we look more examples, let’s revise some of the basic differentiation
formulas to avoid confusion.
d
dx
(c) = 0 Constant Differentiation
d
dx
(xn
) = nxn−1
Power Rule
d
dx
(ax
) = ax
ln x Derivative of an Exponential Function
d
dx
(xx
) = xx
(1 + ln x) Logarithmic Differentiation
It’s very useful to keep on eye on the theorems we use to avoid costly mistakes
since a lot of things look similar but are mathematically different.

Example (Differentiate f(x) = (5 − 3x2
)7
√
6x2 + 8x − 12)
Solution
ln f(x) = ln
h
(5 − 3x2
)7
√
6x2 + 8x − 12
i
= ln(5 − 3x2
)7
+ ln(6x2
+ 8x − 12)
1
2
ln f(x) = 7 ln(5 − 3x2
) +
1
2
ln(6x2
+ 8x − 12)
f′
(x)
f(x)
= 7

−6x
5 − 3x2

+
1
2

12x + 8
6x2 + 8x − 12

f′
(x) = f(x)

−42x
5 − 3x2
+
6x + 4
6x2 + 8x − 12

∴ f′
(x) = (5 − 3x2
)7
√
6x2 + 8x − 12

−42x
5 − 3x2
+
6x + 4
6x2 + 8x − 12


Example (Differentiate y = (2x − e8x
)sin 2x
)
Solution
ln y = ln 2x − e8x
sin 2x
= sin 2x ln(2x − e8x
)
y′
y
= (2 cos 2x) · ln(2x − e8x
) + sin 2x ·

2 − 8e8x
2x − e8x

y′
=

2 cos 2x ln(2x − e8x
) +
sin 2x(2 − 8e8x
)
2x − e8x

× y
∴ y′
=

2 cos 2x ln(2x − e8x
) +
sin 2x(2 − 8e8x
)
2x − e8x

(2x − e8x
)sin 2x


Example (Differentiate y =
√
5x+8
3
√
1 − 9 cos 4x
4
√
t2 + 10t
.)
ln y = ln
√
5x + 8 3
√
1 − 9 cos 4x
4
√
t2 + 10t
!
= ln(5x + 8)
1
2 + ln(1 − 9 cos 4x)
1
3 − ln(t2
+ 10t)
1
4
ln y =
1
2
ln(5x + 8) +
1
3
ln(1 − 9 cos 4x) −
1
4
ln(t2
+ 10t)
y′
y
=
1
2
·
5
5x + 8
+
1

31 ·

3612
sin 4x
1 − 9 cos 4x
−
1

42 ·

2t + 10t+5
t2 + 10t
y′
=

5
2(5x + 8)
+
12 sin 4x
1 − 9 cos 4x
+
t + 5
2(t2 + 10t)

× y
∴ y′
=

5
2(5x + 8)
+
12 sin 4x
1 − 9 cos 4x
+
t + 5
2(t2 + 10t)
√
5x + 8 3
√
1 − 9 cos 4x
4
√
t2 + 10t
#

Exponential Logarithmic Functions Logarithmic Integration
Logarithmic Integration
So far, we have seen that the derivative of ln x is 1
x
. And let’s we want to
integrate 1
x
using the power rule;
R
xn
dx = xn+1
n+1
+ c.
Firstly we will write 1
x
as x−1
using the rules of indices. If we apply the power
rule here, Z
x−1
dx =
x−1+1
−1 + 1
=
x0
0
=
1
0
which is undefined.
Now, knowing the fact that the integral is an antiderivative of a given function,
we can happily conclude from Equation 4 that ln x is an antiderivative of 1
x
on
the interval (−∞, 0) ∪ (0, ∞) that doesn’t contain 0. Therefore;
Z
1
u
du = ln|u|+C

Example (Apply Equation 4 to evaluate
e
R
1
1
x
dx)
Solution
We will firstly find the indefinite integral then handle the limits later. Thus,
Z
1
x
= ln|x|+C
Bringing the limits in,
e
Z
1
1
x
dx = [ln x]e
1 = ln e − ln 1 = 1 − 0 = 1
from rules of logarithms, we know that ln e = loge e = 1 and ln 1 = 0.

Example (Evaluate
R 3x2
x3+5
dx.)
Solution
We will be using Integration by Substitution. The idea is to eliminate the a
certain part of the expression by substituting a derivative of another part of the
given expression. Here, let u = x3
+ 5, after differentiating
du
dx
= 3x2
. Making
dx a subject of the formula, we get dx = du
3x2 . Hence we will substitute u and dx;
Z

3x2
u
·
du

3x2 =
Z
1
u
du = ln u + C
And substitution u back,yields
Z
3x2
x3 + 5
dx = ln|x3
+ 5|+C

Example (Evaluate
R
tan x dx)
Solution
Here, we will first have to apply the trig identity, tan x = sin x
cos x
.
Z
tan x dx =
Z
sin x
cos x
dx
u = cos x and du = − sin x dx
Thus; Z
sin x
cos x
dx = −
Z
du
u
= − ln|u|+C
Substituting u back; Z
tan x dx = − ln|cos x|+C

Example (Integrate
R dx
x ln x
.)
Solution
Firstly, we need to make a choice of u that will eliminate dx as shown
Let u = ln x; Then du =
1
x
dx
Integrating the resulting expression yields the following
Z
dx
x ln x
=
Z
du
u
= ln|u|+C
Substituting u back, Z
dx
x ln x
= ln|ln x|+C

Trigonometric Functions
Trigonometric Functions

Trigonometric Functions Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric Functions
Mathematically, the six basic trigonometric functions do not have inverses
because their graphs repeat periodically and do not pass the horizontal line test.
However, inverses of trigonometric functions are widely used in finding angles
that are associated with specific ratios. These inverses usually give angles which
are in the range −π
2
≤ θ ≤ π
2
. For example, consider
sin x =
1
2
; x = sin−1 1
2
= 30◦
When differentiating these functions, careful attention has to be taken to avoid
confusing them with normal function as their behaviour is not even the same
with the inverses of other functions.

To derive the derivatives of inverse functions, we will use the formula for finding
derivatives of general functions. Thus if f(x) and g(x) are inverses of each other,
then
g′
(x) =
1
f′(g(x))
(6)
And always remember from equation 1 that, two functions f(x) and g(x) are
inverses if and only if f(g(x)) = g(f(x)) = x.
In this section, we will derive the derivatives of the inverses for all the six
trigonometric functions.
Before we start diving into the derivations, it might be important to note that
the inverses are written in two forms, that is, sin−1
x can also be written as
arcsin x.

The Derivative of y = sin−1
x
From previous knowledge, we know that if y = sin−1
x, then, x = sin y. Since,
these two functions are inverses of each other, their derivative will follow
Equation 6. Thus,
d
dx
(sin−1
x) =
dy
dx
=
1
cos y
=
1
cos(sin−1
x)
To solve for the equivalence of cos(sin−1
x), we can use a lot of methods. One of
the methods is transforming the question into a triangle. Since, cos y = x, we
can also write this as cos y = x
1
. Using SOHCAHTOA, x is the opposite side
and 1 is the hypotenuse to angle y of a right-angled triangle. Consider the
figure below,

√
1 − x2
x
1
sin−1
x
cos(sin−1
x) =
√
1 − x2
Here, the third side,
√
1 − x2 has been found by using the Pythagoras
theorem.And therefore,
d
dx
(arcsin x) =
1
√
1 − x2
(7)

Another method to find the same result is to use implicit differentiation and
trigonometric identities.
Given y = sin−1
x then, x = sin y
And if you differentiate x = sin y implicitly,
1 = cos y ×
dy
dx
and
dy
dx
=
1
cos y
Using cos2
y + sin2
y = 1, we know that cos2
y = 1 − sin2
y and
cos y =
p
1 − sin2
y. Thus
dy
dx
=
1
p
1 − sin2
y

Since sin y = x, sin2
y = x2
,
p
1 − sin2
y =
√
1 − x2. And after we substitute,
d
dx
(arcsin x) =
1
√
1 − x2

The Derivative of y = cos−1
x
Using the inverse derivative equation,
d
dx
(cos−1
x) =
dy
dx
= −
1
sin y
= −
1
sin(cos−1 x)
For cos y = x, let x be the adjacent side and 1 be the hypotenuse. Consider the
figure below
x
√
1 − x2
1
cos−1
x

sin(cos−1
x) =
√
1 − x2
Therefore, if we substitute this third side we get the final derivative of arccos as
follows.
d
dx
(arccos x) = −
1
√
1 − x2
Using identities;
Given y = cos−1
x, then x = cos y
1 = − sin y ×
dy
dx
and
dy
dx
= −
1
sin y

But sin2
y + cos2
y = 1, hence, sin y =
p
1 − cos2 y and sin y =
√
1 − x2 because
it has been shown from the question that when y = cos−1
x, x = cos y.
−
1
sin y
= −
1
p
1 − cos2 y
= −
1
√
1 − x2
∴
d
dx
(arccos) = −
1
√
1 − x2
(8)

The Derivative of y = tan−1
x
Firstly, we will use the inverse functions derivative equation.
d
dx
(tan−1
x) =
dy
dx
=
1
sec2 y
=
1
sec2(tan−1
x)
After drawing a triangle in the figure below to represent tan y = x
1
,
1
x
√
1 + x2
tan−1
x

Since sec x = 1
cos x
, using SOHCAHTOA, sec x =
Hypotenuse
Adjacent
. In this case,
sec(tan−1
x) =
√
1 + x2
1
=
√
1 + x2
Therefore; sec2
(tan−1
x) = (
√
1 + x2)2
= 1 + x2
∴
d
dx
(arctan x) =
1
1 + x2
(9)
You can try to use the identities to verify this as well. In addition, the
derivations of the reciprocal functions, Secant, Cosecant and Cotangent
have been intentionally left out for your practice.

Derivatives of Inverse Trig Functions in Summary
d
dx
(sin−1
u) =
1
√
1 − u2
·
du
dx
d
dx
(cos−1
u) = −
1
√
1 − u2
·
du
dx
d
dx
(tan−1
u) =
1
1 + u2
·
du
dx
d
dx
(cot−1
u) = −
1
1 + u2
·
du
dx
d
dx
(sec−1
u) =
1
|u|
√
u2 − 1
·
du
dx
d
dx
(csc−1
u) = −
1
|u|
√
u2 − 1
·
du
dx
Derivatives (10)

Example (Differentiate y = sin−1
(x3
) with respect to x)
Solution
We will use chain rule to differentiate this. Thus;
Let u = x3
;
du
dx
= 3x2
This leaves us with y = sin−1
u.
dy
dx
=
dy
du
×
du
dx
=
1
√
1 − u2
× 3x2
=
3x2
p
1 − (x3)2
∴
d
dx
(sin−1
x3
) =
3x2
√
1 − x6

Example (Differentiate y = sec−1
(ex
) with respect to x)
Solution
Let u = ex du
dx
= ex
Since
d
dx
(sec−1
u) =
1
|u|
√
u2 − 1
·
du
dx
; Then,
d
dx
(sec−1
(ex
)) =
1
|

ex
|
p
(ex)2 − 1
·

ex
∴
d
dx
(sec−1
(ex
)) =
1
√
e2x − 1

Example (Find
dy
dx
if y = x2
(sin−1
x)3
)
Solution
Let u = sin−1
x
du
dx
=
1
√
1 = −x2
This leaves us with y = x2
u3
. Using Product rule;
dy
dx
= 2xu3
+ 3x2
u2 du
dx
Substitute everything back;
∴
dy
dx
= 2x(sin−1
x)3
+ 3x2
(sin−1
x)2 1
√
1 − x2

Integrals Involving Inverse Trigonometric Functions
The derivatives on Equation 10, are very useful in integrating some expressions.
Looking at the derivatives correctly, pairs of the derivatives are similar.
As such, only the positive derivatives are used for the integrals and the negative
ones can be solved using the same positive derivatives.
Below are the integral functions we will use.
Z
1
√
1 − u2
du = sin−1
u + C
Z
1
1 + u2
·
du
dx
du = tan−1
u + C
Z
1
|u|
√
u2 − 1
du = sec−1
u + C
(11)

Example (Evaluate
R ex
√
1−e2x dx)
Solution
We will integration by substitution
Let u = ex
;
du
dx
= ex
. Then dx =
du
ex
Z
ex
√
1 − e2x
dx =
Z

ex
√
1 − u2
×
du

ex =
Z
1
√
1 − u2
du = sin−1
u + C
Substituting u back gives
Z
ex
√
1 − e2x
dx = sin−1
(ex
) + C

Example (Evaluate
R 1
1+3x2 dx.)
Solution
The only challenging part that we can encounter is finding the u-substitution.
But the idea is to make sure that the terms in the radical are squared. For
example, 3x2
as a perfect square will be (
√
3x)2
. Thus
u =
√
3 x du =
√
3 dx and dx =
du
√
3
Z
1
1 + 3x2
dx =
Z
1
1 + u2
du
√
3
=
1
√
3
Z
1
1 + u2
du =
1
√
3
tan−1
u + C
Substituting u back,
Z
1
1 + 3x2
dx =
1
√
3
tan−1
√
3 x

+ C

Example (Evaluate
R dx
a2+x2 where a ̸= 0 is a constant.)
Solution
If you look at Equation 11 closely, you will note that, the first term of the
denominators is always 1. Therefore, we need to change this term to 1 by
factorising a2
on the denominator.
Z
1
a2 1 + x2
a2
dx =
Z
1
a2

1 + x
a
2
dx
Let u =
x
a
; du =
1
a
dx and dx = a du
After we substitute these,
Z
1
a
2 (1 + u2)

a du =
1
a
Z
1
1 + u2
du

Example (Evaluate
R dx
a2+x2 where a ̸= 0 is a constant.)
Continued...
After the constant, 1
a

, has been taken out, we can integrate the remain part.
1
a
Z
1
1 + u2
du =
1
a
tan−1
u + C
And here, we will substitute u back into the original expression.
∴
Z
dx
a2 + x2
=
1
a
tan−1
x
a

+ C

This process can be used to integrate slightly complicated expressions. Similar
processes done for the other integrals give a set usable equations we can use to
simplify the work.
Here are the equations to use whenever the constant a ̸= 0 is not equal to 1;
Z
du
√
a2 − u2
= sin−1 u
a
+ C
Z
du
a2 + u2
=
1
a
tan−1
u
a

+ C
Z
du
u
√
u2 − a2
=
1
a
sec−1 u
a
+ C
(12)

Example (Evaluate
R dx
√
2−x2 )
Solution
There two approaches we can use to solve this question. Firstly, we can simply
use the formulas on equation 12, find the necessary substitutions and integrate,
or we can go through the integration process step by step.
Firstly, let’s use the formula. Thus let’s make both the terms inside the radical
squared. Z
dx
√
2 − x2
=
Z
dx
q √
2
2
− x2
Here, a =
√
2 and u = x; Since
Z
du
√
a2 − u2
= sin−1 u
a
+ C,
Z
dx
√
2 − x2
= sin−1

x
√
2

+ C

Example (Evaluate
R dx
√
9−x2 )
Solution
Here let’s use the step by step approach. (But you can use any method). Start
with factorising 9 to remain with 1 on the first term of the radical.
Z
dx
√
9 − x2
=
Z
dx
q
9 1 − x2
9

We will take out 9 from the root and using the rules of surds, it will be 3 out.
Then we will find u and proceed with substitution.
=
Z
dx
3
q
1 − x
3
2
u =
x
3
and du =
dx
3

Example (Evaluate
R dx
√
9−x2 )
Continued//....
Here, we will substitute any necessary expressions and proceed to integrate.
=
Z

3du

3
√
1 − u2
=
Z
du
√
1 − u2
= u + C
Since we know u = x
3
we will substitute this back..
∴
Z
dx
√
9 − x2
= sin−1
x
3

+ C

Example (Evaluate
R dy
y
√
5y2−3
)
Solution
Factorise 3 and take it out of the radical.
Z
dy
y
p
5y2 − 3
=
Z
dy
y
r
3

5y2
3
− 1
=
Z
dy
√
3y
r√
5y
√
3
2
− 1
Let u =
√
5y
√
3
, then dy =
√
3 du
√
5
Let’s do the substitutions here:
=
Z

√
3du
√
5

√
3 y
p
(u2 − 1)
=
Z
du
√
5 y
p
(u2 − 1)

Example (Evaluate
R dy
y
√
5y2−3
)
Continued/....
Here, we have eliminated y from the inside of the radical and we are only left
with y to eliminate.
Since u =
√
5y
√
3
, then, y =
√
3 u
√
5
, Thus:
Z
du

√
5 ·
√
3 u

√
5
p
(u2 − 1)
=
Z
du
√
3 u
p
(u2 − 1)
=
1
√
3
Z
du
u
p
(u2 − 1)
∴
Z
dy
y
p
5y2 − 3
=
1
√
3
Z
du
u
p
(u2 − 1)
=
1
√
3
sec−1
√
5 y
√
3
+ C

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