Math 2351 Tutorial notes

Differentiation

Definition of Differentiation of function

Def 1 Let f : I and c I . Then We define

f ( x) f (c)
f (c): lim (if limit exists)
xc x c

Thm 2 Let f : I and c I . Then

f is differentiable at c f is continuous at c

Power Rule

Thm 4 Let f ( x) xn , n . Then

f ( x) nxn1

Ex 5 Evaluate the derivative of f ( x) x3 .

Solution: f ( x) 3x31 3x2

Thm 7 Let f ( x) c where c is a constant. Then f ( x) 0 .

Ex 8 Differentiate f ( x) 3x2 x 2 .

Solution: Applying the rules,

f ( x) 3x 2 x 2 3 x 2 x 2
32 x 1 0 6 x 1
1
 Math 2351 Tutorial notes

Rules of Differentiation
In calculating derivatives, we can use the following results to speed up the calculation.

Thm 6 Let f , g : I and c I . If f, g are differentiable at c. Then we have

(a) , f is differentiable at c

and f (c) f (c)

(b) f g is differentiable at c

and f g (c) f (c) g (c)

(c) fg is differentiable at c

and fg (c) f (c) g (c) f (c) g (c)

f
(d) If g (c) 0 then is differentiable at c
g
f f (c) g (c) f (c) g (c)
and (c) 2
g g (c)

Remarks: By using these rules, we can simplify the calculation in finding the
derivatives. For example, we can use the power rule and the above rules to
differentiate any polynomial.

Ex 9
Differentiate f ( x) 3x2 x 1 x3 2 .
Solution: Applying the rules,


f ( x) 3x 2 x 1 x 3 2 3x 2 x 1 x3 2
6 x 1 x 2 3x x 1 3x
3 2 2

15x 4 4 x3 3x 2 12 x 2
 Math 2351 Tutorial notes

Ex 10 Differentiate f ( x) x .
x 12

Solution: Applying the rules,

x x2 1 x x 2 1 1 x2
f ( x)
x x
2 2
2
1 2
1

Derivatives of some special functions

Sometimes, we need to be able to differentiate the trigonometric functions, exponential
functions and logarithmic functions.

(a) sin x cos x (g) e x e x

(b) cos x sin x (h) a x ln a a x

(c) tan x sec2 x (i) ln x 1x

(d) csc x csc x cot x (j) loga x 1
for
ln a x
(e) sec x sec x tan x
a 0 & a 1
(f) cot x csc2 x

Ex 11 Differentiate f ( x) x sin x .
ln x
Solution: Using the results above,

x sin x ln x x sin x ln x
f ( x)
ln x
2

sin x x cos x ln x x sin x 1x

ln x
2

(ln x 1)sinx x cos x ln x

ln x
2
 Math 2351 Tutorial notes

The Chain Rule

Thm 12 (Chain Rule)

Let f : I1 , g : I 2 where f ( I1) I 2 . Let c I1 . Then
f is differentiable at c and g is differentiable at f (c)
g f is differentiable at c and g f (c) g f (c) f (c)

Remarks: There are two different versions of Chain Rule, but they are all equivalent.
(1) d g f ( x) g f ( x) f ( x)
dx
(2) dy dy du
dx du dx

99
Ex 13 Differentiate f ( x) x 2 1 .

Solution: Applying Chain Rule,

x2 1 99 x2 1 2 x 198x x2 1
98 98 98
f ( x) 99 x2 1

Ex 14 Given f (u) sin u and u( x) cos( x) . Find df .

dx

df df du d sin u d cosx
Solution: cos u sin x cos cos( x) sin x
dx du dx du dx
 Math 2351 Tutorial notes

Integration
Fundamental Theorem of Calculus

Thm 22 (Fundamental Thm. of Calculus)

Let f :[a, b] be a continuous function.
x
(1st part): Let F ( x) : f (t )dt .
a

Then F ( x) f ( x) x (a, b)
(2nd part): G is an antiderivative of f on [a, b]
b
a f ( x)dx G(b) G(a)

Remark: Or simply write,

d x
(1st part): f (t )dt f ( x)
dx a
x
(2nd part): a G(t)dt G( x) G(a)

b
a x dx .
2
Ex 23 Evaluate
3
Solution: Let G( x) x . Then G( x) x2 . By Fundamental Theorem of Calculus,
3
b b
a x2dx G(t )dt G(b) G(a )
a

b3 a3 1 3 3
b a
3 3 3

0 cos xdx .
Ex 24 Evaluate 2

2 cos xdx 2 G (t )dt
Solution: Let G( x) sin x . Then G( x) cos x .
0 0

G(b) G(a ) sin sin(0) 1

2
 Math 2351 Tutorial notes

Integrals of Basic Functions

Here is a list of commonly used integration formulas, including trigonometric,
logarithmic and exponential types.
x n1
x dx C
n
Thm 25 (a)
n 1
1
(b) x dx ln x C
e dx e C
x x
(c)
ax
a dx C
x
(d)
ln a
(e) ln xdx x ln x x C
(f) sin xdx cos x C
(g) cos xdx sin x C
(h) tan xdx ln sec x C
(i) csc xdx ln csc x cot x C
(j) sec xdx ln sec x tan x C
(k) cot xdx ln sin x C

Integration by Substitution

Thm 26 (Definite Integration by Substitution)

Let g :[a, b] has continuous derivative on [a, b] .
Let f : g [a, b] be a continuous function. Then

a f g( x) g( x)dx g (a) f (u)du

b g (b)
where u g ( x) .

Remark: We have similar theorem for indefinite Integrals:

f g( x) g( x)dx f (u)du where u g ( x)

 Math 2351 Tutorial notes

(2 x 1)
99
Ex 27 Evaluate dx .

Solution: Let u 2 x 1. Then du 2 (or du 2dx )

dx
1 u99 2dx

2
99 99
(2 x 1) dx u dx
100
1 u99du u C
2 200

Integrating by Parts
Thm 28 (Integrating by Parts)
Let f , g :[a, b] be continuously differentiable functions. Then
b b b
a f g fg a fg
a

Remarks: (a) The condition continuously differentiable is to make sure that f

and g are integrable.
(b) Derivatives may not be integrable.

Ex 29 Evaluate ln xdx .
Solution: Integrating by parts,

ln xdx ln x x dx ln x x x ln x dx

ln x x x 1 dx ln x x x C
x

Complex Number
Def 30 A complex number is an expression of the form z x iy ,

in which x and y are real numbers and i is the imaginary unit: i 2 1.

We define: Re( z) x Im( z) y

 Math 2351 Tutorial notes

Def 31 Let z . Let z x iy . Then we define

(complex conjugate of z) z x iy

(modulus of z) z x2 y 2
y
(argument of z) arg(z) is such that tan
x
If is taken between ( , ] then it is called the principal argument.

Prop 32 Let z1, z2 . Let z1 x1 iy1 and z2 x2 iy2 . Then

(a) z1 z2 x1 x2 & y1 y2
(b) z1 z2 ( x1 x2 ) i( y1 y2 )
(c) z1z2 ( x1 iy1)( x2 iy2 ) ( x1x2 y1 y2 ) i( y1x2 x1 y2 )
x1 iy1 x2 iy2 x1 x2 y1 y2 i y1 x2 x1 y2
(d) z1 x1 iy1

z2 x2 iy2 x2 iy2 x2 iy2 x2 2 y2 2

Thm 33 Let z1, z2 . Let z1 x1 iy1 and z2 x2 iy2 . Then

z z
(b) z1 z2 z1 z2 (c) z1z2 z1 z2
2
(a) zz z (d) 1 1
z z
2 2

n
Def 30 We define exp( z) z z .
n0 n!

Remarks: (a) This series converges for all complex z. (can be proved by ratio test)
(b) exp( x) ex x
(c) exp( z) has the similar properties of e x .

Def 30 We define cos Re exp(i ) & sin Im exp(i ) for all real .
Or, ei cos i sin
Remark: This definition matches the geometric meaning of the trigo. functions.