First principle, power rule, derivative of constant term, product rule, quotient rule, chain rule, derivatives of trigonometric functions and their inverses, derivatives of exponential functions and natural logarithmic functions, implicit differentiation, parametric differentiation, L'Hopital's rule

1. SMA 2101: CALCULUS I
c
⃝Francis O. Ochieng
francokech@gmail.com
YouTube: Prof. Francis Oketch∗
Department of Pure and Applied Mathematics
Jomo Kenyatta University of Agriculture and Technology
Course content
• Functions: definition, domain, range, codomain, composition (or composite), inverse.
• Limits, continuity and differentiability of a function.
• Differentiation by first principle and by rule for xn (integral and fractional n).
• Other techniques of differentiation, i.e., sums, products, quotients, chain rule; their applications
to algebraic, trigonometric, logarithmic, exponential, and inverse trigonometric functions all of
a single variable.
• Implicit and parametric differentiation.
• Applications of differentiation to: rates of change, small changes, stationary points, equations of
tangents and normal lines, kinematics, and economics and financial models (cost, revenue and
profit).
• Introduction to integration and its applications to area and volume.
References
[1] Calculus: Early Transcendentals (8th Edition) by James Stewart
[2] Calculus with Analytic Geometry by Roland E. Larson, Robert P. Hostetler and Bruce H. Edwards;
5th edition
[3] Calculus and Analytical Geometry (9th edition) by George B. Thomas and Ross L. Finney
[4] Advanced Engineering Mathematics (10th ed.) by Erwin Kreyszig
[5] Calculus by Larson Hostellem
Lecture 4
1 Derivatives of functions
Definition 1.1 (First principle). The derivative of a function f(x) denoted by f′(x) or
df
dx
is the rate
of change of f with respect to x, and is given by
.
.
f′
(x) = lim
h→0
[
f(x + h) − f(x)
h
]
,
for all x for which this limit exists.
∗
https://www.youtube.com/channel/UC7wd3x6 08GNoUbLxmaC4vA
1

2. c
⃝Francis Oketch
The process of finding the derivative f′(x) is called differentiation of f(x). The above relation is
called first principle of differentiation or differentiation by the definition or differentiation of first kind.
Geometrically, consider the curve y = f(x) and let ∆x = h.
Here, ∆y = f(x + h) − f(x). So, gradient of the secant
line through points A and B is
∆y
∆x
=
f(x + h) − f(x)
h
.
Taking limit as h → 0 yields
dy
dx
= lim
h→0
f(x + h) − f(x)
h
= f′
(x)
Therefore,
.
.
dy
dx
= f′
(x)
Example(s):
(a) Use first principle of differentiation to find the derivative of the function f(x) = x2.
Solution
Given f(x) = x2, we have f(x + h) = (x + h)2. By the first principle of differentiation, we have
f′
(x) = lim
h→0
f(x + h) − f(x)
h
= lim
h→0
(x + h)2 − x2
h
= lim
h→0
x2 + 2hx + h2 − x2
h
= lim
h→0
2hx + h2
h
= lim
h→0
(2x + h)
D.S
= (2x + 0)
= 2x
(b) Use first principle of differentiation to find the derivative of the following functions: (i) f(x) =
1
x
and (ii) f(x) =
√
x.
Solution
i) Given f(x) =
1
x
, we have f(x + h) =
1
x + h
. By the first principle of differentiation, we
have
f′
(x) = lim
h→0
f(x + h) − f(x)
h
= lim
h→0
(
1
x+h − 1
x
)
h
= lim
h→0
x − (x + h)
hx(x + h)
= lim
h→0
−h
hx(x + h)
= lim
h→0
−1
x(x + h)
D.S
=
−1
x(x + 0)
= −
1
x2
ii) Given f(x) =
√
x, we have f(x + h) =
√
x + h. By the first principle of differentiation, we
have
f′
(x) = lim
h→0
f(x + h) − f(x)
h
= lim
h→0
√
x + h −
√
x
h
= lim
h→0
(
√
x + h −
√
x)(
√
x + h +
√
x)
h(
√
x + h +
√
x)
= lim
h→0
x + h − x
h(
√
x + h +
√
x)
= lim
h→0
h
h(
√
x + h +
√
x)
= lim
h→0
1
√
x + h +
√
x
D.S
=
1
√
x + 0 +
√
x
=
1
2
√
x
2

3. 1.1 Basic differentiation rules c
⃝Francis Oketch
Exercise:
(a) Use first principle of differentiation to find the derivative of the following functions.
i) f(x) = −x3 + 3x2 + 4 [ans: f′(x) = −3x2 + 6x]
ii) f(x) =
3x
1 − 5x
[ans: f′(x) =
3
(1 − 5x)2
]
iii) f(x) =
−7 + 5x
−3 − 2x
[ans: f′(x) =
−29
(−3 − 2x)2
]
iv) f(x) =
√
6x + 2 − 5 [ans: f′(x) =
3
√
6x + 2
]
v) f(x) =
1
√
x + 2
[ans: f′(x) =
−1
2
√
x(
√
x + 2)2
]
1.1 Basic differentiation rules
The derivative of a constant
If f(x) = c (a constant) for all x, then f′(x) = 0 for all x. That is, .
.
dc
dx
= f′
(x) = 0 .
Proof. Given f(x) = c ⇒ f(x + h) = c. Thus, from the first principle, we have
f′
(x) = lim
h→0
f(x + h) − f(x)
h
= lim
h→0
c − c
h
= lim
h→0
0
h
= 0
The power rule
If f(x) = xn for n ∈ R, then .
.
f′
(x) = nxn−1
. That is, bring down the power and reduce the
power by one.
Proof. Given f(x) = xn ⇒ f(x+h) = (x+h)n = xn +nxn−1h+
n(n − 1)
2!
xn−2h2 +· · ·+hn.
Thus, from the first principle, we have
f′
(x) = lim
h→0
f(x + h) − f(x)
h
= lim
h→0
(
xn + nxn−1h +
n(n − 1)
2!
xn−2h2 + · · · + hn − xn
)
h
= lim
h→0
(
nxn−1h +
n(n − 1)
2!
xn−2h2 + · · · + hn
)
h
= lim
h→0
(
nxn−1
+
n(n − 1)
2!
xn−2
h + · · · + hn−1
)
D.S
=
(
nxn−1
+
n(n − 1)
2!
xn−2
(0) + · · · + (0)n−1
)
= nxn−1
For example,
i) If f(x) = 6x5, then f′(x) = 30x4.
ii) If f(x) = x10, then f′(x) = 10x9.
The derivative of a linear combination
If f(x) and g(x) are differentiable functions of x and a and b are constants, then
.
.
d
dx
[af(x) + bg(x)] = af′
(x) + bg′
(x)
3

4. 1.1 Basic differentiation rules c
⃝Francis Oketch
Proof. Let y(x) = af(x) + bg(x). Thus, from the first principle, we have
dy
dx
= lim
h→0
y(x + h) − y(x)
h
= lim
h→0
[af(x + h) + bg(x + h)] − [af(x) + bg(x)]
h
= lim
h→0
a[f(x + h) − f(x)] + b[g(x + h) − g(x)]
h
= a lim
h→0
f(x + h) − f(x)
h
+ b lim
h→0
g(x + h) − g(x)
h
= af′
(x) + bg′
(x)
For example,
i) If y = 24x + 8x5, then
dy
dx
= 24 + 40x4.
ii) If y = 7x3 − 9x2 + 4x + 2, then
dy
dx
= 21x2 − 18x + 4.
The product rule
If u(x) and v(x) are differentiable functions of x, then the product u(x)v(x) is also a differentiable
function of x, and
.
.
d
dx
[u(x)v(x)] = u′
(x)v(x) + u(x)v′
(x)
Proof. Let y(x) = u(x)v(x). Thus, from the first principle, we have
dy
dx
= lim
h→0
y(x + h) − y(x)
h
= lim
h→0
u(x + h)v(x + h) − u(x)v(x)
h
= lim
h→0
u(x + h)v(x + h) − u(x)v(x + h) + u(x)v(x + h) − u(x)v(x)
h
= lim
h→0
[u(x + h) − u(x)] v(x + h) + u(x) [v(x + h) − v(x)]
h
= lim
h→0
[u(x + h) − u(x)] v(x + h)
h
+ lim
h→0
u(x) [v(x + h) − v(x)]
h
=
[
lim
h→0
u(x + h) − u(x)
h
] [
lim
h→0
v(x + h)
]
+ u(x) lim
h→0
v(x + h) − v(x)
h
= u′
(x)v(x) + u(x)v′
(x)
→ Note: the product rule says that the derivative of the product of two functions is formed by
multiplying the derivative of each function by the other function and then adding the results.
In general, suppose y = u1(x)u2(x) · · · un(x), then
.
.
dy
dx
= u′
1(x)u2(x) · · · un(x) + u1(x)u′
2(x) · · · un(x) + · · · + u1(x)u2(x) · · · u′
n(x)
Example(s):
(a) Find the derivative of f(x) = (1 − 5x2)(6x2 − 4x + 1).
Solution
Let f(x) = uv, where u = 1 − 5x2 and v = 6x2 − 4x + 1. Differentiating yields u′ = −10x
and v′ = 12x − 4. Therefore,
f′
(x) = u′
v + uv′
= (−10x)(6x2
− 4x + 1) + (1 − 5x2
)(12x − 4)
= −60x3
+ 40x2
− 10x + 12x − 4 − 60x3
+ 20x2
= −120x3
+ 60x2
+ 2x − 4
4

5. 1.1 Basic differentiation rules c
⃝Francis Oketch
(b) Find the derivative of y = (x − 2)(x2 + 6)(x4 + 1).
Solution
Let y = uvw, where u = x − 2, v = x2 + 6 and w = x4 + 1. Differentiating yields
u′ = 1, v′ = 2x and w′ = 4x3. Therefore,
dy
dx
= u′
vw + uv′
w + uvw′
= (1)(x2
+ 6)(x4
+ 1) + (x − 2)(2x)(x4
+ 1) + (x − 2)(x2
+ 6)(4x3
)
=
(
x6
+ x2
+ 6x4
+ 6
)
+ 2x
(
x5
+ x − 2x4
+ 2
)
+ 4x3
(
x3
+ 6x − 2x2
− 12
)
= x6
+ x2
+ 6x4
+ 6 + 2x6
+ 2x2
− 4x5
+ 4x + 4x6
+ 24x4
− 8x5
− 48x3
= 7x6
− 12x5
+ 30x4
− 48x3
+ 3x2
+ 4x + 6
The quotient rule
If u(x) and v(x) are differentiable functions of x, then the quotient
u(x)
v(x)
(where v(x) ̸= 0) is also
a differentiable function of x, and
.
.
d
dx
[
u(x)
v(x)
]
=
u′(x)v(x) − u(x)v′(x)
[v(x)]2
Proof. Let y(x) =
u(x)
v(x)
. Thus, from the first principle, we have
dy
dx
= lim
h→0
y(x + h) − y(x)
h
= lim
h→0
u(x + h)
v(x + h)
−
u(x)
v(x)
h
= lim
h→0
u(x + h)v(x) − u(x)v(x + h)
hv(x)v(x + h)
= lim
h→0
u(x + h)v(x) − u(x)v(x) + u(x)v(x) − u(x)v(x + h)
hv(x)v(x + h)
= lim
h→0
[u(x + h) − u(x)] v(x) − u(x) [v(x + h) − v(x)]
hv(x)v(x + h)
=
lim
h→0
[u(x + h) − u(x)]
h
v(x) − u(x) lim
h→0
[v(x + h) − v(x)]
h
lim
h→0
v(x)v(x + h)
=
u′(x)v(x) − u(x)v′(x)
[v(x)]2
Example(s):
(a) Differentiate y =
2x2 + 1
x2 − 1
.
Solution
Let y =
u
v
, where u = 2x2 + 1 and v = x2 − 1. Differentiating yields u′ = 4x and v′ = 2x.
Therefore,
dy
dx
=
vu′ − uv′
v2
=
(x2 − 1)(4x) − (2x2 + 1)(2x)
(x2 − 1)2
=
4x3 − 4x − 4x3 − 2x
(x2 − 1)2
=
−6x
(x2 − 1)2
(b) Differentiate y =
x3
x − 1
.
Solution
5

6. 1.1 Basic differentiation rules c
⃝Francis Oketch
Let y =
u
v
, where u = x3 and v = x − 1. Differentiating yields u′ = 3x2 and v′ = 1.
Therefore,
dy
dx
=
vu′ − uv′
v2
=
(x − 1)(3x2) − (x3)(1)
(x − 1)2
=
3x3 − 3x2 − x3
(x − 1)2
=
2x3 − 3x2
(x − 1)2
The chain rule
Suppose that y is a differentiable function of u and u is a differentiable function of x (i.e.,
y = y(u) and u = u(x)), then y is a (differentiable) function of x by extension (i.e., y = y(u(x)))
and
.
.
dy
dx
=
dy
du
·
du
dx
→ Note: chain rule is used when we want to differentiate a function of another function.
Example(s):
(a) Differentiate with respect to x the function y = (3x + 4)4.
Solution
Let y = u4, where u = 3x + 4. Differentiating yields
dy
du
= 4u3 and
du
dx
= 3. Therefore,
chain rule yields
dy
dx
=
dy
du
·
du
dx
= (4u3
)(3) = 12u3
= 12(3x + 4)3
(b) Differentiate with respect to x the function y = (x2 + 3x)7.
Solution
Let y = u7, where u = x2 +3x. Differentiating yields
dy
du
= 7u6 and
du
dx
= 2x+3. Therefore,
chain rule yields
dy
dx
=
dy
du
·
du
dx
= (7u6
)(2x + 3) = 7(x2
+ 3x)6
(2x + 3)
(c) Find
dy
dx
if y = (1 − 3x2)5.
Solution
Let y = u5, where u = 1 − 3x2. Differentiating yields
dy
du
= 5u4 and
du
dx
= −6x. Therefore,
chain rule yields
dy
dx
=
dy
du
·
du
dx
= (5u4
)(−6x) = −30x(u4
) = −30x(1 − 3x2
)4
(d) Find
dy
dx
if y =
(
1 + 2x
1 + x
)2
.
Solution
Let y = u2, where u =
1 + 2x
1 + x
. Differentiating yields
dy
du
= 2u and
du
dx
=
(1 + x)(2) − (1 + 2x)(1)
(1 + x)2
=
1
(1 + x)2
. Therefore, chain rule yields
dy
dx
=
dy
du
·
du
dx
= (2u)
[
1
(1 + x)2
]
= 2
[
1 + 2x
1 + x
] [
1
(1 + x)2
]
=
2(1 + 2x)
(1 + x)3
6

7. 1.2 Derivative of trigonometric functions c
⃝Francis Oketch
(e) Differentiate with respect to x the function y =
√
1 +
√
1 +
√
1 + x.
Solution
Let u =
√
1 + x, v =
√
1 + u, and w =
√
1 + v. Then, y = w. Differentiating yields
dy
dw
= 1,
du
dx
=
1
2(1 + x)
1
2
,
dv
du
=
1
2(1 + u)
1
2
, and
dw
dv
=
1
2(1 + v)
1
2
. Therefore, chain rule yields
dy
dx
=
dy
dw
·
dw
dv
·
dv
du
·
du
dx
= (1)
[
1
2(1 + v)
1
2
] [
1
2(1 + u)
1
2
] [
1
2(1 + x)
1
2
]
=
1
8(1 + v)
1
2 (1 + u)
1
2 (1 + x)
1
2
=
1
8
(√
1 + v
) (√
1 + u
) (√
1 + x
)
=
1
8
(√
1 +
√
1 +
√
1 + x
) (√
1 +
√
1 + x
)
(√
1 + x
)
→ Note: (Direct chain rule)
Consider the function y = [f(x)]n
. Then, direct chain rule yields
.
.
dy
dx
= n [f(x)]n−1
· f′
(x)
For example, if y = (1 − 3x2)5 then DCR yields
dy
dx
= 5(1 − 3x2)4(0 − 6x) = −30x(1 − 3x2)4.
Exercise:
(a) Use chain rule to differentiate the following functions
i) y = (3x2 + 5)3
ii) y = (3x3 + 5x)2
iii) y = (7x2 − 4)
1
3
iv) y = (6x2 − 4x)−2
v) y = (3x2 − 5)−2
3
vi) y = (1 + x4 − 2x3)4(1 − 4x2)3
(b) Find
dy
dx
when y =
√
1 + x
1 − x
. [ans:
dy
dx
=
1
(1 + x)
1
2 (1 − x)
3
2
]
(c) Differentiate with respect to x the function y =
√
x +
√
x +
√
x.
Lecture 4
1.2 Derivative of trigonometric functions
Derivative of sin x and cos x
.
.
d
dx
[sin x] = cos x and
d
dx
[cos x] = − sin x
Proof. i) Sine
7

8. 1.2 Derivative of trigonometric functions c
⃝Francis Oketch
Let f(x) = sin x. Thus, from the first principle of differentiation and using the trigonometric
identity sin(A + B) = sin A cos B + sin B cos A, we have
f′
(x) = lim
h→0
f(x + h) − f(x)
h
= lim
h→0
sin(x + h) − sin x
h
= lim
h→0
sin x cos h + sin h cos x − sin x
h
= lim
h→0
− sin x(1 − cos h) + sin h cos x
h
= − sin x
[
lim
h→0
(1 − cos h)
h
]
+ cos x
[
lim
h→0
sin h
h
]
= (− sin x)(0) + (cos x)(1)
= cos x
ii) Cosine
Similarly, let f(x) = cos x. Thus, from the first principle of differentiation and using the
trigonometric identity cos(A + B) = cos A cos B − sin A sin B, we have
f′
(x) = lim
h→0
f(x + h) − f(x)
h
= lim
h→0
cos(x + h) − cos x
h
= lim
h→0
cos x cos h − sin x sin h − cos x
h
= lim
h→0
− cos x(1 − cos h) − sin x sin h
h
= − cos x
[
lim
h→0
(1 − cos h)
h
]
− sin x
[
lim
h→0
sin h
h
]
= (− cos x)(0) − (sin x)(1)
= − sin x
Example(s):
(a) Differentiate the following functions wrt x: (i) y = sin(3x + 2), (ii) y = cos3 x, (iii) y =
sin(x2), (iv) y = x sin(x), (v) y =
sin x
x
, and (vi) y = cos2(3x).
Solution
i) Given that y = sin(3x + 2). Let y = sin(u), where u = 3x + 2. Differentiating yields
dy
du
= cos u and
du
dx
= 3. Therefore, chain rule yields
dy
dx
=
dy
du
·
du
dx
= (cos u)(3) = 3 cos(3x + 2)
ii) Given that y = cos3 x. Let y = u3, where u = cos x. Differentiating yields
dy
du
= 3u2
and
du
dx
= − sin x. Therefore, chain rule yields
dy
dx
=
dy
du
·
du
dx
= (3u2
)(− sin x) = −3 sin x cos2
x
iii) Given that y = sin(x2). Let y = sin(u), where u = x2. Differentiating yields
dy
du
= cos u
and
du
dx
= 2x. Therefore, chain rule yields
dy
dx
=
dy
du
·
du
dx
= (cos u)(2x) = 2x cos(x2
)
iv) Given that y = x sin(x). Let y = uv, where u = x and v = sin(x). Differentiating
yields u′ = 1 and v′ = cos x. Therefore, product rule yields
dy
dx
= uv′
+ vu′
= (x)(cos x) + (sin x)(1) = x cos x + sin x
8

9. 1.2 Derivative of trigonometric functions c
⃝Francis Oketch
v) Given that y =
sin x
x
. Let y =
u
v
, where u = sin x and v = x. Differentiating yields
u′ = cos x and v′ = 1. Therefore, quotient rule yields
dy
dx
=
vu′ − uv′
v2
=
(x)(cos x) − (sin x)(1)
x2
=
x cos x − sin x
x2
vi) Given that y = cos2(3x). Let y = u2, where u = cos(3x). Differentiating yields
dy
du
= 2u and
du
dx
= −3 sin(3x). Therefore, chain rule yields
dy
dx
=
dy
du
·
du
dx
= (2u)[−3 sin(3x)] = −6 cos(3x) sin(3x)
(b) If y =
√
1 + sin x, show that
dy
dx
=
1
2
√
1 − sin x.
Solution
Let y = u
1
2 , where u = 1 + sin x. Differentiating yields
dy
du
=
1
2
u−1
2 and
du
dx
= cos x.
Therefore, chain rule yields
dy
dx
=
dy
du
·
du
dx
=
(
1
2
u− 1
2
)
(cos x) =
cos x
2
√
u
=
cos x
2
√
1 + sin x
=
1
2
cos x
√
1 − sin x
(
√
1 + sin x)(
√
1 − sin x)
=
1
2
cos x
√
1 − sin x
√
1 − sin2 x
=
1
2
cos x
√
1 − sin x
√
cos2 x
=
1
2
√
1 − sin x
(c) Find
dy
dx
if y = sin(cos x).
Solution
Let y = sin u, where u = cos x. Differentiating yields
dy
du
= cos u and
du
dx
= − sin x.
Therefore, chain rule yields
dy
dx
=
dy
du
·
du
dx
= (cos u)(− sin x) = − cos(cos x) sin x
In general,
.
.
d
dx
[sin (f(x))] = f′
(x) cos (f(x)) and
d
dx
[cos (f(x))] = −f′
(x) sin (f(x))
Exercise:
(a) Find
dy
dx
if y = sin(
√
x).
(b) If y =
√
1 + sin x
1 − sin x
, show that
dy
dx
=
1
1 − sin x
.
(c) If m is a positive integer, find the differential coefficients with respect to x of (i) sinm x and
(ii) sin(xm).
(d) Differentiate the following functions with respect to x: (i) y = sin 3x, (ii) y = cos(x2),
(iii) y =
√
sin 2x, (iv) y = 4 sin2(x
2 ), (v) y = sin x cos 2x, (vi) y =
cos 2x
sin 3x
, and (vii)
y = 2 cos x + 2x sin x − x2 cos x.
Derivative of tan x, cot x, sec x and cosec x
9

10. 1.2 Derivative of trigonometric functions c
⃝Francis Oketch
I Tangent
Let y = tan x =
sin x
cos x
. Differentiating using quotient rule yields
dy
dx
=
(cos x)(cos x) − (sin x)(− sin x)
cos2 x
=
cos2 x + sin2 x
cos2 x
=
1
cos2 x
= sec2
x
Therefore, .
.
d
dx
[tan x] = sec2
x .
II Cotangent
Let y = cot x =
cos x
sin x
. Differentiating using quotient rule yields
dy
dx
=
(sin x)(− sin x) − (cos x)(cos x)
sin2 x
=
−
[
sin2 x + cos2 x
]
sin2 x
=
−1
sin2 x
= −cosec2
x
Therefore, .
.
d
dx
[cot x] = −cosec2
x .
III Secant
Let y = sec x =
1
cos x
. Differentiating using quotient rule yields
dy
dx
=
(cos x)(0) − (1)(− sin x)
cos2 x
=
sin x
cos2 x
=
1
cos x
·
sin x
cos x
= sec x tan x
Therefore, .
.
d
dx
[sec x] = sec x tan x .
IV Cosecant
Let y = cosec x =
1
sin x
. Differentiating using quotient rule yields
dy
dx
=
(sin x)(0) − (1)(cos x)
sin2 x
=
− cos x
sin2 x
=
−1
sin x
·
cos x
sin x
= −cosec x cot x
Therefore, .
.
d
dx
[cosec x] = −cosec x cot x .
In summary, we have
f(x) f′(x)
sin x cos x
cos x − sin x
tan x sec2 x
cosec x −cosec x cot x
sec x sec x tan x
cot x −cosec2x
Example(s):
(a) Differentiate the following functions with respect to x: (i) y = tan 2x, (ii) y = cot 3x, (iii)
y = 3 sec 2x, (iv) y = sec x tan x, (v) y = x2 cot x, and (vi) y =
x
tan x
.
Solution
10

11. 1.2 Derivative of trigonometric functions c
⃝Francis Oketch
i) Given that y = tan 2x. Let y = tan u, where u = 2x. Differentiating yields
dy
du
= sec2 u
and
du
dx
= 2. Therefore, chain rule yields
dy
dx
=
dy
du
·
du
dx
= (sec2
u)(2) = 2 sec2
(2x)
ii) Given that y = cot 3x. Let y = cot u, where u = 3x. Differentiating yields
dy
du
=
−cosec2u and
du
dx
= 3. Therefore, chain rule yields
dy
dx
=
dy
du
·
du
dx
= (−cosec2
u)(3) = −3cosec2
(3x)
iii) Given that y = 3 sec 2x. Let y = 3 sec u, where u = 2x. Differentiating yields
dy
du
=
3 sec u tan u and
du
dx
= 2. Therefore, chain rule yields
dy
dx
=
dy
du
·
du
dx
= (3 sec u tan u)(2) = 6 sec(2x) tan(2x)
iv) Given that y = sec x tan x. Let y = uv, where u = sec x and v = tan x. Differentiating
yields u′ = sec x tan x and v′ = sec2 x. Therefore, product rule yields
dy
dx
= uv′
+ vu′
= (sec x)(sec2
x) + (tan x)(sec x tan x) = sec3
x + sec x tan2
x
v) Given that y = x2 cot x. Let y = uv, where u = x2 and v = cot x. Differentiating
yields u′ = 2x and v′ = −cosec2x. Therefore, product rule yields
dy
dx
= uv′
+ vu′
= (x2
)(−cosec2
x) + (cot x)(2x) = 2x cot x − x2
cosec2
x
vi) Given that y =
x
tan x
. Let y =
u
v
, where u = x and v = tan x. Differentiating yields
u′ = 1 and v′ = sec2 x. Therefore, quotient rule yields
dy
dx
=
vu′ − uv′
v2
=
(tan x)(1) − (x)(sec2 x)
tan2 x
=
tan x − x sec2 x
tan2 x
In general,
.
.
d
dx
[tan (f(x))] = f′
(x) sec2
(f(x)) and
d
dx
[cot (f(x))] = −f′
(x)cosec2
(f(x))
.
.
d
dx
[sec (f(x))] = f′
(x) sec (f(x)) tan (f(x)) and
d
dx
[cosec (f(x))] = −f′
(x)cosec (f(x)) cot (f(x))
Exercise:
(a) Differentiate the following functions with respect to x: (i) y = sec2 2x, (ii) y = 3 sec x tan x,
(iii) y = −cosec2
(
1
2 x
)
, and (iv) y =
sec x
x
.
(b) If y = (tan x + sec x)m
, where m is a positive integer. Show that
dy
dx
= my sec x.
Lecture 5
11

12. 1.3 Derivative of exponential functions c
⃝Francis Oketch
1.3 Derivative of exponential functions
A standard exponential function of x takes the form y = ex or y = exp(x). The number e ≈ 2.7183 is
called Euler’s constant. To differentiate the standard exponential function, use the rule
.
.
d
dx
[ex
] = ex
Proof. Given y(x) = ex ⇒ y(x + h) = ex+h = exeh. Thus, by the first principle of differentiation,
we have
y′
(x) = lim
h→0
[
y(x + h) − y(x)
h
]
= lim
h→0
[
exeh − ex
h
]
= ex
lim
h→0
[
eh − 1
h
]
From the table, we have
h 0.0001 0.001 0.01 0.1 -0.01 -0.001
eh − 1
h
1.0005 1.05
Hence, lim
h→0
[
eh − 1
h
]
= 1
Therefore, y′(x) = ex(1) = ex.
In general, suppose y = ef(x). Let y = eu, where u = f(x). Differentiating yields
dy
du
= eu and
du
dx
= f′(x). By chain rule, we have
dy
dx
=
dy
du
·
du
dx
= eu · f′(x) = f′(x)ef(x). Therefore,
.
.
d
dx
[
ef(x)
]
= f′
(x)ef(x)
In other words, to differentiate y = ef(x), multiply the exponential function by the derivative of the
exponent.
Example(s):
(a) Find
dy
dx
given (i) y = e−6x and (ii) y = ex2
Solution
i) Given y = e−6x. Let u = −6x ⇒ y = eu. Differentiating we get
du
dx
= −6 and
dy
du
= eu.
Hence, chain rule yields
dy
dx
=
dy
du
·
du
dx
= −6eu = −6e−6x. Therefore,
dy
dx
= −6e−6x.
ii) Given y = ex2
. Let u = x2 ⇒ y = eu. Differentiating we get
du
dx
= 2x and
dy
du
= eu.
Hence, chain rule yields
dy
dx
=
dy
du
·
du
dx
= 2xeu = 2xex2
. Therefore,
dy
dx
= 2xex2
.
(b) Differentiate the following functions with respect to x: (i) y = 2e−3x + e4x and (ii) y = 2esin 3θ.
Solution
Using direct chain rule, we have
i)
dy
dx
= 2
d
dx
[
e−3x
]
+
d
dx
[
e4x
]
= 2
[
e−3x d
dx
(−3x)
]
+
[
e4x d
dx
(4x)
]
= −6e−3x + 4e4x.
ii)
dy
dθ
= 2
d
dθ
[
esin 3θ
]
= 2
[
esin 3θ d
dθ
(sin 3θ)
]
= 2esin 3θ(3 cos 3θ) = 6esin 3θ cos 3θ.
(c) If y = e−2x cos 4x, find
dy
dx
.
Solution
12

13. 1.4 Derivative of natural logarithmic functions c
⃝Francis Oketch
Product rule yields
dy
dx
= e−2x d
dx
[cos 4x] + cos 4x
d
dx
[
e−2x
]
= −4e−2x
sin 4x − 2e−2x
cos 4x
Exercise:
1. Find
dy
dx
given:
i) y =
√
xe
√
x + e
√
x2+1
ii) y =
e2x
1 + x2e3x
iii) y =
(
6 + e3x cos 4x
)4
iv) y = esin 5x + 2x2ecos 3x
v) y =
(
e3x2+6x
)
cos (ex + e−x)
vi) y = etan(4+sin 3x) +
1
e2x + e5x
+ 6.
1.4 Derivative of natural logarithmic functions
A natural logarithmic function of x is logarithm of x to base e, i.e., loge x = ln x. Thus in index form,
x can be rewritten as x = eln x. Similarly, y = eln y, a = eln a, etc. Suppose y = ln x. Then,
.
.
d
dx
[ln x] =
1
x
Proof. Given y = ln x. Taking exponential on both sides yields ey = x. Differentiating both sides
with respect to x yields ey dy
dx
= 1 ⇒
dy
dx
=
1
ey
=
1
x
. Therefore,
dy
dx
=
1
x
.
In general, suppose y = ln[f(x)]. Then, ey = f(x). Differentiating both sides with respect to x yields
ey dy
dx
= f′
(x) ⇒
dy
dx
=
f′(x)
ey
=
f′(x)
f(x)
Therefore,
.
.
d
dx
{ln[f(x)]} =
f′(x)
f(x)
In other words, to differentiate y = ln[f(x)], multiply the reciprocal of the parameter by the derivative
of the parameter.
Example(s):
(a) Find
dy
dx
given: (i) y = ln(x2), (ii) y = ln(cos 2x), (iii) y = ln
( √
x2 + 1
3
√
x3 + 1
)
.
Solution
i) Given y = ln(x2). Let u = x2 ⇒ y = ln(u). Differentiating yields
du
dx
= 2x and
dy
du
=
1
u
=
1
x2
. Hence, chain rule yields
dy
dx
=
dy
du
·
du
dx
=
2x
x2
=
2
x
. Therefore,
dy
dx
=
2
x
13

14. 1.4 Derivative of natural logarithmic functions c
⃝Francis Oketch
ii) Let y = ln(u) where u = cos 2x. Differentiating yields
dy
du
=
1
u
and
du
dx
= −2 sin 2x. Hence,
chain rule yields
dy
dx
=
dy
du
·
du
dx
=
−2 sin 2x
u
=
−2 sin 2x
cos 2x
= −2 tan 2x. Therefore,
dy
dx
= −2 tan 2x
iii) Given y = ln
( √
x2 + 1
3
√
x3 + 1
)
⇒ y = ln
[
(x2 + 1)1/2
(x3 + 1)1/3
]
=
1
2
ln
(
x2 + 1
)
−
1
3
ln
(
x3 + 1
)
.
Thus, y =
1
2
ln
(
x2 + 1
)
−
1
3
ln
(
x3 + 1
)
. Differentiating both sides with respect to x yields
dy
dx
=
1
2
d
dx
[
ln
(
x2
+ 1
)]
−
1
3
d
dx
[
ln
(
x3
+ 1
)]
=
1
2 (x2 + 1)
d
dx
(
x2
+ 1
)
−
1
3 (x3 + 1)
d
dx
(
x3
+ 1
)
=
1
2 (x2 + 1)
(2x) −
1
3 (x3 + 1)
(
3x2
)
=
x
(x2 + 1)
−
x2
(x3 + 1)
=
x(x3 + 1) − x2(x2 + 1)
(x2 + 1)(x3 + 1)
=
x4 + x − x4 − x2)
(x2 + 1)(x3 + 1)
=
x − x2
(x2 + 1)(x3 + 1)
(b) Find
dy
dx
given that y = sin (ln 2x).
Solution
Let y = sin u, where u = ln 2x. Thus, chain rule yields
dy
dx
=
dy
du
·
du
dx
= (cos u)
[
1
2x
·
d
dx
(2x)
]
= (cos u)
[
1
x
]
=
cos u
x
=
cos(ln 2x)
x
(c) Find
dy
dx
given y =
√
1 + x
1 − x
.
Solution
Given y =
√
1 + x
1 − x
. Taking natural logarithm on both sides yields
ln y = ln
(
1 + x
1 − x
)1/2
=
1
2
ln
(
1 + x
1 − x
)
=
1
2
[ln(1 + x) − ln(1 − x)]
Thus, ln y =
1
2
[ln(1 + x) − ln(1 − x)]. Differentiating with respect to x yields
d
dx
[ln y] =
1
2
{
d
dx
[ln(1 + x)] −
d
dx
[ln(1 − x)]
}
⇒
1
y
dy
dx
=
1
2
{
1
(1 + x)
d
dx
(1 + x) −
1
(1 − x)
d
dx
(1 − x)
}
=
1
2
{
1
1 + x
+
1
1 − x
}
=
1
1 − x2
Therefore,
dy
dx
=
1
1 − x2
y =
1
1 − x2
√
1 + x
1 − x
=
1
(1 + x)(1 − x)
·
(1 + x)1/2
(1 − x)1/2
=
1
(1 + x)1/2(1 − x)3/2
Exercise:
(a) i) If y = (
√
x − 1)ex ln x, find
dy
dx
. [ans:
dy
dx
=
ex
2x
√
x − 1
{
(2x2 − x) ln x + 2x − 2
}
]
14

15. 1.5 Implicit differentiation c
⃝Francis Oketch
ii) Find the gradient of the curve y = ln(
√
1 + sin 2x) at the point where x =
π
2
. [ans: = -1]
(b) Find
dy
dx
given:
i) y = ln
√
2x + 6
ii) y =
√
x ln(
√
x)
iii) y =
1 + 2x2 ln 3x
1 +
√
sec(ln 2x)
iv) y =
1
1 + 2 ln 46x
−
1
sin(ln(15x2))
v) y =
(
2x2 + ln
√
x
)6
(1 + 2x sec 2x)3
vi) y = cot
(
ln 2x + e3x
)
Lecture 6
1.5 Implicit differentiation
An implicit function is a function where the dependent variable y is not expressed explicitly in terms
of the independent variable x (i.e., a function where y is not the subject of the formula). To find
dy
dx
,
follow these steps:
i) Differentiate x normally
ii) Apply direct chain rule in differentiating y
iii) Collect like terms and make
dy
dx
the subject
Example(s):
1. Find
dy
dx
given x4 + y5 = 125.
Solution
Differentiating the given equation implicitly with respect to x, we get
4x3
+ 5y4 dy
dx
= 0 ⇒
dy
dx
= −
4x3
5y4
2. Find
dy
dx
given y + xy + y2 = 2.
Solution
Differentiating the given equation implicitly with respect to x, we get
dy
dx
+ x
dy
dx
+ (1)y + 2y
dy
dx
= 0 ⇒
dy
dx
=
−y
1 + x + 2y
3. Find
dy
dx
when y3 − 3x2y + 2x3 = 0.
Solution
Differentiating the given equation implicitly with respect to x, we get
3y2 dy
dx
− 3x2 dy
dx
− 6xy + 6x2
= 0 ⇒ (3y2
− 3x2
)
dy
dx
= 6xy − 6x2
15

16. 1.6 Differentiation of other forms of exponential functions c
⃝Francis Oketch
Therefore,
dy
dx
=
6x(y − x)
3(y2 − x2)
=
6x(y − x)
3(y + x)(y − x)
=
2x
y + x
4. If y2 − 2y
√
1 + x2 + x2 = 0, show that
dy
dx
=
x
√
1 + x2
.
Solution
Differentiating the given equation implicitly with respect to x, we get
2y
dy
dx
−2
√
1 + x2
dy
dx
−2y
[
1
2
(1 + x2
)−1
2
]
(2x)+2x = 0 ⇒
(
2y − 2
√
1 + x2
) dy
dx
=
2xy
√
1 + x2
−2x
Therefore,
dy
dx
=
2xy
√
1 + x2
− 2x
2y − 2
√
1 + x2
=
2x
(
y −
√
1 + x2
)
2
√
1 + x2
(
y −
√
1 + x2
) =
x
√
1 + x2
Exercise:
1. Find
dy
dx
given:
(a) xy3 − 2x2y2 + x4 = 1
(b) x2 sin y − y cos x = 10x3
(c) x cos y − y2 sin x = 2
(d) exy2
= 10
(
x2 + y2
)
(e) ln(x2 +
√
y) = sin(xy2)
(f) tan y sin x = cos(xy)
(g) cos(x + y) sin(x − y) = 20x2
(h) y2ex2y =
30
√
xy
(i) ln
(
x + y
x2y
)
= 10x2
(j) cos(xy) = 4x2 + y2
2. Find
dy
dx
and
d2y
dx2
given that 2x2y3 + x4 = 6y2 + 2x.
1.6 Differentiation of other forms of exponential functions
Consider exponential functions of the form
I: A constant raised to a function (e.g., y = 10tan 3x)
In this case, introduce natural logarithm on both sides first. On the left differentiate implicitly
and on the right hand side differentiate normally.
Example(s):
(a) Find
dy
dx
given: (i) y = ax2
where a is a constant, (ii) y = 3−x2+6x+10, and (iii) y = 4sin 5x.
Solution
16

17. 1.6 Differentiation of other forms of exponential functions c
⃝Francis Oketch
i) Given y = ax2
. Taking natural logarithm on both sides yields
ln y = x2
ln a
Differentiating with respect to x yields
d
dx
[ln y] =
d
dx
[x2
ln a] ⇒
1
y
dy
dx
= 2x ln a ⇒
dy
dx
= 2xy ln a
Therefore,
dy
dx
= 2xax2
ln a.
ii) Given y = 3−x2+6x+10. Taking natural logarithm on both sides yields
ln y = (−x2
+ 6x + 10) ln 3
Differentiating with respect to x yields
1
y
dy
dx
= (−2x + 6) ln 3 ⇒
dy
dx
= (−2x + 6)y ln 3
Therefore,
dy
dx
= (−2x + 6)3−x2+6x+10 ln 3.
iii) Given y = 4sin 5x. Taking natural logarithm on both sides yields
ln y = sin 5x ln 4
Differentiating with respect to x yields
1
y
dy
dx
= 5 cos 5x ln 4 ⇒
dy
dx
= (5 cos 5x)y ln 4
Therefore,
dy
dx
= (5 cos 5x)4sin 5x ln 4.
Exercise:
i) Find
dy
dx
given y = ax + bx, where a and b are constants. [ans:
dy
dx
= ax ln a + bx ln b]
II: A function raised to a function (e.g., y = xtan 3x)
In this case, introduce natural logarithm on both sides first. On the left differentiate implicitly
and on the right hand side differentiate using product rule.
Example(s):
(a) Find
dy
dx
given: (i) y = xx, (ii) y = (tan x)x
, (iii) y = (sin 4x)x2
, and (iv) y =
(
ex2
)x
.
Solution
i) Given y = xx. Taking natural logarithm on both sides yields
ln y = x ln x
Differentiating with respect to x yields
d
dx
[ln y] =
d
dx
[x ln x]
⇒
1
y
dy
dx
= x
d
dx
[ln x] + ln x
d
dx
(x) = x ·
1
x
+ ln x
= 1 + ln x
Therefore,
dy
dx
= y(1 + ln x) = xx(1 + ln x).
17

18. 1.6 Differentiation of other forms of exponential functions c
⃝Francis Oketch
ii) Given y = (tan x)x
. Taking natural logarithm on both sides yields
ln y = x ln(tan x)
Differentiating with respect to x yields
d
dx
[ln y] =
d
dx
[x ln(tan x)]
⇒
1
y
dy
dx
= x
d
dx
[ln(tan x)] + ln(tan x)
d
dx
(x) = x ·
sec2 x
tan x
+ ln(tan x)
=
x sec2 x
tan x
+ ln(tan x)
Therefore,
dy
dx
= y
(
x sec2 x
tan x
+ ln(tan x)
)
= (tan x)x
(
x sec2 x
tan x
+ ln(tan x)
)
.
iii) Given y = (sin 4x)x2
. Taking natural logarithm on both sides yields
ln y = x2
ln(sin 4x)
Differentiating with respect to x yields
d
dx
[ln y] =
d
dx
[x2
ln(sin 4x)]
⇒
1
y
dy
dx
= x2 d
dx
[ln(sin 4x)] + ln(sin 4x)
d
dx
(x2
) = x2
·
4 cos 4x
sin 4x
+ 2x ln(sin 4x)
= 4x2
cot 4x + 2x ln(sin 4x)
Therefore,
dy
dx
= y
[
4x2 cot 4x + 2x ln(sin 4x)
]
= (sin 4x)x2 [
4x2 cot 4x + 2x ln(sin 4x)
]
.
Exercise:
(a) Differentiate the following functions with respect to x.
i) y = xcos x. [ans:
dy
dx
= xcos x
(
− sin x ln x +
cos x
x
)
]
ii) y = (sin x)x. [ans:
dy
dx
= (sin x)x (ln(sin x) + x cot x)]
iii) y = (x + x2)x. [ans:
dy
dx
= (x + x2)x
(
ln(x + x2) +
1 + 2x
1 + x
)
]
iv) yx = x. [ans:
dy
dx
=
y
x
(
1
x
− ln y
)
]
v) xy = sin x. [ans:
dy
dx
=
1
ln x
(
cot x −
y
x
)
]
vi) ysin x =
√
x. [ans:
dy
dx
=
y
sin x
(
1
2x
− cot x ln y
)
]
(b) Find
dy
dx
given:
i) y = (cos 3x)sin 3x
ii) y = ax + sin (2x)
iii) y = (cos x)x
+ 2x2
iv) y = sin
(
a2x + 3x
)
III: Derivatives of other logarithmic functions e.g., y = log2(3x2 + 1)
To find
dy
dx
, first convert the logarithm to index notation then introduce natural logarithm on
both sides.
Example(s):
18

19. 1.7 Inverse trigonometric functions c
⃝Francis Oketch
(a) Find
dy
dx
given: (i) y = log2(3x2 + 1), (ii) y = logsin x x, and (iii) y = logx(cos 3x)
Solution
i) Given y = log2(3x2 + 1). In index notation, we have 2y = 3x2 + 1. Taking natural
logarithm on both sides yields
y ln 2 = ln(3x2
+ 1)
Differentiating with respect to x yields
d
dx
[y ln 2] =
d
dx
[ln(3x2
+ 1)] ⇒
dy
dx
ln 2 =
6x
3x2 + 1
Therefore,
dy
dx
=
6x
(3x2 + 1) ln 2
.
ii) Given y = logsin x x. In index notation, we have (sin x)y
= x. Taking natural logarithm
on both sides yields
y ln(sin x) = ln x
Differentiating with respect to x yields
y
cos x
sin x
+ ln(sin x)
dy
dx
=
1
x
⇒ y cot x + ln(sin x)
dy
dx
=
1
x
Therefore,
dy
dx
=
(
1
x
− y cot x
)
1
ln(sin x)
.
iii) Given y = logx(cos 3x). In index notation, we have xy = cos 3x. Taking natural
logarithm on both sides yields
y ln x = ln(cos 3x)
Differentiating with respect to x yields
y
x
+ ln(x)
dy
dx
=
−3 sin 3x
cos 3x
⇒
y
x
+ ln(x)
dy
dx
= −3 tan 3x
Therefore,
dy
dx
=
(
−
y
x
− 3 tan 3x
)
1
ln x
.
Lecture 7
1.7 Inverse trigonometric functions
y = sin−1
x = arcsin x, y = cos−1
x = arccos x, y = tan−1
x = arctan x
y = cosec−1
x = arccosec x, y = sec−1
x = arcsec x, y = cot−1
x = arccot x.
→ Note: sin−1 x ̸=
1
sin x
, cos−1 x ̸=
1
cos x
, etc.
1.7.1 Derivative of inverse trigonometric functions
To find
dy
dx
, follow these steps
i) Introduce the trigonometric function corresponding to the given inverse on both sides of the
given equation
ii) Differentiate implicitly on the left hand side and differentiate normally on the right hand side.
iii) Make
dy
dx
the subject.
19

20. 1.7 Inverse trigonometric functions c
⃝Francis Oketch
To find a suitable form of the trigonometric function in the denominator,
put the given equation in place of y, or
replace the denominator by either making use of trigonometric identities or draw a right angled
triangle and find the missing side using Pythagoras theorem, as follows:
For example,
I Let y = sin−1 x ⇒ sin y = x. Differentiating both sides with respect to x yields
cos y
dy
dx
= 1 ⇒
dy
dx
=
1
cos y
Formula 1: putting y = sin−1 x in place of y yields
dy
dx
=
1
cos
(
sin−1 x
)
Formula 2:
using the identity cos2 y + sin2 y = 1 ⇒ cos y =
√
1 − sin2 y. Thus,
dy
dx
=
1
√
1 − sin2 y
. Putting sin y = x yields
dy
dx
=
1
√
1 − x2
, or
using the right angled triangle
.
.
√
1 − x2
.
x
.
1
.
y
From sin y = x, we have opposite is x and hypotenuse is 1. From
Pythagoras theorem, the adjacent is given by
√
1 − x2. From the
diagram, cos y =
√
1 − x2. Hence,
dy
dx
=
1
√
1 − x2
Therefore, .
.
d
dx
[
sin−1
x
]
=
1
√
1 − x2 .
II Let y = cos−1 x ⇒ cos y = x. Differentiating both sides with respect to x yields
− sin y
dy
dx
= 1 ⇒
dy
dx
=
−1
sin y
=
−1
√
1 − cos2 y
=
−1
√
1 − x2
Therefore, .
.
d
dx
[
cos−1
x
]
=
−1
√
1 − x2 .
III Let y = tan−1 x ⇒ tan y = x. Differentiating both sides with respect to x yields
sec2
y
dy
dx
= 1 ⇒
dy
dx
=
1
sec2 y
=
1
1 + tan2 y
=
1
1 + x2
Therefore, .
.
d
dx
[
tan−1
x
]
=
1
1 + x2 .
IV Let y = cot−1 x ⇒ cot y = x. Differentiating both sides with respect to x yields
−cosec2
y
dy
dx
= 1 ⇒
dy
dx
=
−1
cosec2y
=
−1
1 + cot2 y
=
−1
1 + x2
Therefore, .
.
d
dx
[
cot−1
x
]
=
−1
1 + x2 .
V Let y = sec−1 x ⇒ sec y = x. Differentiating both sides with respect to x yields
sec y tan y
dy
dx
= 1 ⇒
dy
dx
=
1
sec y tan y
=
1
sec y
(√
sec2 y − 1
) =
1
x
√
x2 − 1
Therefore, .
.
d
dx
[
sec−1
x
]
=
1
x
√
x2 − 1
.
20

21. 1.7 Inverse trigonometric functions c
⃝Francis Oketch
VI Let y = cosec−1x ⇒ cosec y = x. Differentiating both sides with respect to x yields
−cosec y cot y
dy
dx
= 1 ⇒
dy
dx
=
−1
cosec y cot y
=
−1
cosec y
(√
cosec2y − 1
) =
−1
x
√
x2 − 1
Therefore, .
.
d
dx
[
cosec−1
x
]
=
−1
x
√
x2 − 1
.
Example(s):
(a) Find
dy
dx
if y = sin−1(2x2 + x + 1).
Solution
Let y = sin−1 u, where u = 2x2 + x + 1. Differentiating yields
dy
du
=
1
√
1 − u2
and
du
dx
= 4x + 1.
Therefore, chain rule yields
dy
dx
=
dy
du
·
du
dx
=
(
1
√
1 − u2
)
(4x + 1) =
4x + 1
√
1 − (2x2 + x + 1)2
(b) Find
dy
dx
if y = cos−1(2x + 1).
Solution
Let y = cos−1 u, where u = 2x+1. Differentiating yields
dy
du
=
−1
√
1 − u2
and
du
dx
= 2. Therefore,
chain rule yields
dy
dx
=
dy
du
·
du
dx
=
(
−1
√
1 − u2
)
(2) =
−2
√
1 − (2x + 1)2
(c) Find
dy
dx
if y = tan−1(cos x + x).
Solution
Let y = tan−1 u, where u = cos x + x. Differentiating yields
dy
du
=
1
1 + u2
and
du
dx
= − sin x + 1.
Therefore, chain rule yields
dy
dx
=
dy
du
·
du
dx
=
(
1
1 + u2
)
(− sin x + 1) =
− sin x + 1
1 + (cos x + x)2
(d) Differentiate y = x sin−1 x with respect to x.
Solution
Let y = uv, where u = x and v = sin−1 x. Differentiating yields u′ = 1 and v′ =
1
√
1 − x2
.
Therefore, product rule yields
dy
dx
= uv′
+ vu′
= (x)
(
1
√
1 − x2
)
+ (sin−1
x)(1) =
x
√
1 − x2
+ sin−1
x
(e) Find
dθ
dt
when (i) θ = cos−1(1 − 2t2) and (ii) θ = sin−1(2t3 − 1).
Solution
21

22. 1.7 Inverse trigonometric functions c
⃝Francis Oketch
i) Given that θ = cos−1(1 − 2t2). Let θ = cos−1 u, where u = 1 − 2t2. Differentiating yields
dθ
du
=
−1
√
1 − u2
and
du
dt
= −4t. Therefore, chain rule yields
dθ
dt
=
dθ
du
·
du
dt
=
(
−1
√
1 − u2
)
(−4t) =
4t
√
1 − (1 − 2t2)2
=
4t
√
1 − 1 + 4t2 − 4t4
=
4t
√
4t2(1 − t2)
=
4t
2t
√
1 − t2
=
2
√
1 − t2
ii) Given that θ = sin−1(2t3 − 1). Let θ = sin−1 u, where u = 2t3 − 1. Differentiating yields
dθ
du
=
1
√
1 − u2
and
du
dt
= 6t2. Therefore, chain rule yields
dθ
dt
=
dθ
du
·
du
dt
=
(
1
√
1 − u2
)
(6t2
) =
6t2
√
1 − (2t3 − 1)2
=
6t2
√
1 − 4t6 + 4t3 − 1
=
4t
√
4t2(t − t4)
=
6t2
2t
√
t − t4
=
3t
√
t − t4
(f) If y =
(
tan−1 x
)2
, prove that
d
dx
{
(1 + x2)
dy
dx
}
=
2
1 + x2
.
Proof. Let y = u2, where u = tan−1 x. Differentiating yields
dy
du
= 2u and
du
dx
=
1
1 + x2
.
Therefore, chain rule yields
dy
dx
=
dy
du
·
du
dx
= (2u)
(
1
1 + x2
)
=
2 tan−1 x
1 + x2
Now,
d
dx
{
(1 + x2
)
dy
dx
}
=
d
dx
{
(1 + x2
)
2 tan−1 x
1 + x2
}
=
d
dx
{
2 tan−1
x
}
=
2
1 + x2
Exercise:
(a) i) If y = sin−1(cos x), show that
dy
dx
= −1.
ii) If y = sin−1(3x − 4x3), show that
√
1 − x2
dy
dx
= 3.
iii) If u = θ2 +
(
sin−1 θ
)2
− 2θ
√
1 − θ2 sin−1 θ, show that
√
1 − θ2
du
dθ
= 4θ2 sin−1 θ.
(b) Find the derivative of the following functions: (i) y = x2
(
sin−1)x
)2
, (ii) y =
√
tan−1 x, and (iii)
y = sin
(
tan−1 x
)
.
(c) Find
dy
dx
given that y = sin−1 √
x. [ans:
dy
dx
=
1
2
√
x − x2
]
(d) Find
dy
dx
given:
i) y = esin−1(3x) +
1
2x + cos−1(4x)
ii) y = 2x + cos−1(4x)
iii) y = ln
(
x2 + 2x
)
iv) y =
sin−1(3x)
23x + sin 3x
22

23. 1.8 Parametric differentiation c
⃝Francis Oketch
v) y = cosec−1(3x)
vi) y = x2cosec−1(4x)
vii) y = xx sin−1(2x)
(e) Trey knows the following values for f(x) and f′(x):
x 5 6 7 8 9
f(x) 8 2 7 9 5
f′(x) -3 1 -8 9 6
i) Let h(x) = ln(5x + 3) tan−1(f(x)). Find h′(8) [ans: h′(8) = 0.5826]
ii) Let m(x) = sec−1(f(x)). Find m′(8) [ans: m′(8) = 1.1118]
(f) Brendan knows the following values for f(x) and f′(x):
x 3 4 5 6 7
f(x) -0.6 0.7 0.7 -0.3 -0.4
f′(x) 10 -9 6 -5 -2
i) Let h(x) = sin−1(f(x)). Find h′(3) [ans: h′(3) = 12.5]
ii) Let m(x) = e5x sin−1(f(x)). Find m′(3) [ans: m′(3) = 3.0345 × 107]
CAT 1
Lecture 8
1.8 Parametric differentiation
If both x and y are defined as functions of another variable (parameter), say t, i.e., x = x(t), y = y(t),
then
.
.
dy
dx
=
(
dy
dt
)
÷
(
dx
dt
)
=
dy/dt
dx/dt
Example(s):
1. Find
dy
dx
, in terms of the parameter t, when (a) x = at2, y = 2at, (b) x = (t + 1)2, y = (t2 − 1),
and (c) x = cos−1(3t), y = sin−1(3t).
Solution
(a)
dx
dt
= 2at,
dy
dt
= 2a. Therefore,
dy
dx
=
dy/dt
dx/dt
=
2a
2at
=
1
t
.
(b)
dx
dt
= 2(t + 1),
dy
dt
= 2t. Therefore,
dy
dx
=
dy/dt
dx/dt
=
2t
2(t + 1)
=
t
t + 1
.
(c) Rewrite as cos x = 3t, sin y = 3t. Differentiating with respect to t yields − sin x
dx
dt
= 3 and
cos y
dy
dt
= 3. Hence,
dx
dt
= −
3
sin x
= −
3
√
1 − 9t2
and
dy
dt
=
3
cos y
=
3
√
1 − 9t2
. Therefore,
dy
dx
=
dy/dt
dx/dt
=
(
3
√
1 − 9t2
)
÷
(
−
3
√
1 − 9t2
)
= −1.
Exercise:
(a) Find
dy
dx
given:
23

24. 1.9 Higher order derivatives c
⃝Francis Oketch
i) x = ln(2t2), y = ln(4 + t2)
ii) x = 2t, y = 2−t
iii) x = tan−1(2t), y = sec−1(2t)
iv) x = t sin(t2), y = t3 cos(t2)
v) x =
t2
1 + t2
, y =
1 − t2
1 + t2
vi) x = et cos 2t, y = e−t sin 2t
vii) x =
2t
1 + 3t
, y = t3 − 4t + 8
viii) x = θ − sin 2θ, y = θ + cos 2θ
ix) x = a cos3 θ, y = b sin3 θ
1.9 Higher order derivatives
Suppose y(x) is an n−times differentiable function of x. Then,
First derivative of y(x) is given by
dy
dx
= y′(x)
Second derivative of y(x) is given by
d
dx
(
dy
dx
)
=
d2y
dx2
= y′′(x)
Third derivative of y(x) is given by
d
dx
(
d2y
dx2
)
=
d3y
dx3
= y′′′(x)
.
.
.
.
.
.
nth derivative of y(x) is given by
d
dx
(
dn−1y
dxn−1
)
=
dny
dxn
= y(n)(x)
Example(s):
1. Given that y = 4x3 − 6x2 − 9x + 1, find
dy
dx
,
d2y
dx2
,
d3y
dx3
, and
d4y
dx4
.
Solution
dy
dx
= 12x2 − 12x − 9,
d2y
dx2
= 24x − 12,
d3y
dx3
= 24,
d4y
dx4
= 0.
2. Find
d2y
dx2
and
d3y
dx3
when (a) y = x10 and (b) y = cos 2x.
Solution
(a)
dy
dx
= 10x9,
d2y
dx2
= 90x8,
d3y
dx3
= 720x7.
(b)
dy
dx
= −2 sin 2x,
d2y
dx2
= −4 cos 2x,
d3y
dx3
= 8 sin 2x.
3. If y =
cos x
x
, prove that
d2y
dx2
+
2
x
dy
dx
+ y = 0.
Proof. Quotient rule yields
dy
dx
=
x
d
dx
[cos x] − cos x
d
dx
[x]
x2
=
−x sin x − cos x
x2
.
d2y
dx2
=
x2 d
dx
[−x sin x − cos x] − (−x sin x − cos x)
d
dx
[x2]
x4
=
x2(−x cos x − sin x + sin x) + (x sin x + cos x)(2x)
x4
=
−x3 cos x + 2x2 sin x + 2x cos x
x4
=
−x2 cos x + 2x sin x + 2 cos x
x3
24

25. c
⃝Francis Oketch
Now,
d2y
dx2
+
2
x
dy
dx
+ y =
−x2 cos x + 2x sin x + 2 cos x
x3
+
2
x
(
−x sin x − cos x
x2
)
+
cos x
x
=
−x2 cos x + 2x sin x + 2 cos x − 2x sin x − 2 cos x + x2 cos x
x3
= 0
4. Given that u and v are functions of x, show that
d3
dx3
(uv) =
d3u
dx3
v + 3
d2u
dx2
dv
dx
+ 3
du
dx
d2v
dx2
+ u
d3v
dx3
.
Proof.
d3
dx3
(uv) =
d
dx
[
d2
dx2
(uv)
]
=
d
dx
[
d
dx
{
d
dx
(uv)
}]
=
d
dx
[
d
dx
{
du
dx
v + u
dv
dx
}]
(product rule)
=
d
dx
[
d2u
dx2
v + 2
du
dx
dv
dx
+ u
d2v
dx2
]
=
d3u
dx3
v +
d2u
dx2
dv
dx
+ 2
d2u
dx2
dv
dx
+ 2
du
dx
d2v
dx2
+
du
dx
d2v
dx2
+ u
d3v
dx3
=
d3u
dx3
v + 3
d2u
dx2
dv
dx
+ 3
du
dx
d2v
dx2
+ u
d3v
dx3
Exercise:
1. Find
d2y
dx2
when (a) y = (1+4x+x2) sin x, (b) y = x tan−1 x, (c) y =
x2
1 + x
, (d) y = (3x−sin 2x)2,
and (e) y = ln(3x3 + 4x − 1) + xex2
.
2. If y =
sin x
x
, find
dy
dx
and
d2y
dx2
. Hence, prove that x2 d2y
dx2
+ 4x
dy
dx
+ (x2 + 2)y = 0.
Lecture 9
2 L’Hôpital’s rule
If on direct substitution we get indeterminate forms such as
0
0
or
±∞
±∞
, then the L’Hôpital’s rule states
that
.
.
lim
x→a
f(x)
g(x)
= lim
x→a
f′(x)
g′(x)
provided the limit exists. Repeat finding derivatives until you get a meaningful result.
Example(s):
1. Evaluate the following using L’Hôpital’s rule
(a) lim
x→2
x2 + x − 6
x − 2
Solution
lim
x→2
x2 + x − 6
x − 2
D.S
=
0
0
(indeterminate)
L′H
= lim
x→2
2x + 1
1
D.S
= 5
25

26. c
⃝Francis Oketch
(b) lim
x→−2
x2 + 3x + 2
2x2 − 8
Solution
lim
x→−2
x2 + 3x + 2
2x2 − 8
D.S
=
0
0
(indeterminate)
L′H
= lim
x→−2
2x + 3
4x
D.S
=
1
8
(c) lim
x→∞
5x3 − 1
4x3 − 2x − 7
.
Solution
lim
x→∞
5x3 − 1
4x3 − 2x − 7
D.S
=
∞
∞
(indeterminate)
L′H
= lim
x→∞
15x2
12x2 − 2
D.S
=
∞
∞
(indeterminate)
L′H
= lim
x→∞
30x
24x
D.S
=
∞
∞
(indeterminate)
L′H
= lim
x→∞
30
24
=
5
4
(d) lim
x→0
x − sin x
x3
Solution
On direct substitution, we have lim
x→0
x − sin x
x3
=
0 − sin 0
03
=
0
0
(indeterminate). So,
lim
x→0
x − sin x
x3
L′H
=
1 − cos x
3x2
D.S
=
0
0
(indeterminate)
L′H
= lim
x→0
sin x
6x
D.S
=
0
0
(indeterminate)
L′H
= lim
x→0
cos x
6
D.S
=
1
6
Exercise:
1. Evaluate
(a) lim
x→0
2 sin x − sin 2x
2ex − 2 − x2
(b) lim
x→∞
7x − 28
x3
(c) lim
x→1
x2 − x − 2
x2 − 1
(d) lim
x→0
sin 7x
4x
. [ans: = 7/4]
(e) lim
x→0+
cot x
ln x2
.
(f) lim
y→π
2
(
cos y
π
2 − y
)
. [ans: = 1]
26

27. c
⃝Francis Oketch
(g) lim
θ→0
1 − cos θ
θ
. [ans: = 0]
(h) lim
θ→0
sin2 θ
θ2
. [ans: = 1]
2. Amy knows the following values for f(x) and g(x): f(−6) = 1, g(−6) = 1, f′(−6) = 6 and
g′(−6) = −5. Assume all functions are continuous, find the following.
(a) lim
x→−6
ln(g(x))
[f(x)]2 − 1
. [ans: = −
5
12
]
(b) lim
x→−6
[g(x) − 1]2
[f(x) − 1]2
. [ans: =
25
36
]
(c) lim
x→−6
g(x)[f(x) − 1]
f(x)[g(x) − 1]
. [ans: = −
6
5
]
27