Differentiation

JALILAH RUBAI
Kolej Matrikulasi Johor

Q Q Q = ( )
P∆ ∆ P∆ ∆
∆ Tangent to curve
P∆ at point P
Line passes
Q through P and Q
P  , = ( , (∆ ))

 ( + ∆ , ( + ∆ ))

 The slope/gradient of the line PQ ( ) is .

 As Q approches P, Δ decreases, approahes 0, the line approaches the tangent line and
approaches the gradient of tangent ( ) at point P.

 Gradient of tangent at any point , , = lim = lim ∆ ()

∆→ ∆→

 Let ∆ = ℎ, = = lim ( ) FIRST PRINCIPLES

→

Example 2: Use differentiation from first principles
to find the gradient function of = .

Example 1: Determine, from the first principles, Solution: + ℎ =
the gradient function for the curve = −
and calculate its value at = 3.

Solution: + ℎ = ( + ℎ) − + ℎ . = ′ = lim
= + 2ℎ + ℎ − − ℎ
→

11 1
= lim + ℎ + 2 − + 2
+ 2ℎ + ℎ − − ℎ − ( − ) ℎ
→

= lim ℎ

→

ℎ(2 + ℎ − 1) = lim 1 + 2 − ( + ℎ + 2)

= lim ℎ → ℎ ( + ℎ + 2)( + 2)

→

= 2 − 1 = lim −1

→ ( + ℎ + 2)( + 2)

3 = 2 3 − 1 = 5 −1 1
= ( + 2)( + 2) = − ( + 2)

− = lim 0 = 0
= lim
ℎ →
→

= lim + ℎ − = lim ℎ = 1

→ ℎ → ℎ

= lim + ℎ − = lim + 2ℎ + ℎ − = lim ℎ 2 + ℎ = 2

→ ℎ → ℎ → ℎ

+ ℎ − + 3 ℎ + 3 ℎ + ℎ − ℎ 3 + 3 ℎ + ℎ
= lim = lim = lim = 3
ℎ ℎ ℎ
→ → →

= lim + ℎ − = lim + 4 ℎ + 6 ℎ + 4 ℎ + ℎ − = lim ℎ 4 + 6 ℎ + 4 ℎ + ℎ = 4

→ ℎ → ℎ → ℎ


= = +

=


= −



=
=

= ( ) → = =

= = ′( )
( ) ± ( ) = ′( ) ± ′( )

= × = ′( )

Example 3: Example 4:
Find the derivative of = 1 − 2 .
Differentiate = 3 + 1 with respect to
Solution: by using the product rule.

= − 2 Solution:

= 2 − 8 = = 3 + 1

= 2 = 6

It often pays to first think about the best
way to approach a problem, before blasting = 6 + 3 + 1 2
away with derivative rules! By rewriting the
function, we can considerably simplify our = 6 + 3 + 6
work. = 12 + 3

Although it is easier to differentiate after
rewriting the function, we have to follow
the instruction!

Example 5: Example 6:

Differentiate Find the derivative of
− 3 + 1
= + 2 − 1
=
with respect to . with respect to by using the quotient rule.

Solution: Solution:

21 = − 3 + 1 =
= + −
= 4 − 6 = 1
= + 2 −

4 − 6 − − 3 + 1 1
= 1 + 2 − − −2 =


22 4 − 6 − + 3 − 1
= 1 − + =

Follow the instruction! 1
= 3 − 3 −
Rewrite the function and then differentiate,
no need to use the quotient rule!

Example 7: , find . Example 8:
Given that = 2 −1 and =
Find the derivative of = 3 − 2
by using the chain rule.

Solution: 1 − Solution: =
= = −
= 2 − 1 = 3 − 2 = 3
= − − 2
= 4 = 3


= × = ×

1 = 3 3
= 4 − − 2
= 9 3 − 2

1 − −1 − 2
= 4

4 − 1 2 + 1 Chain Rule
=

Embedded Chain Rule Example 9: .
Find the derivative of =
When functions are embedded within one another, we need
to take the derivative of the outside part with the original Solution: 1 ( ) − outer
inner part and multiply that by the derivative of the inside = 2 − 3 2 2 − 3 − inner
part. 2
= −2 2 − 3
For example: Differentiate = 1 − 3 . −4

Step 1: Pretend like the inner value of parenthesis simply = 2 − 3
says “ .”
Example 10:
Step 2: Differentiate. Now you have = 4
Find the derivative of = − 3 + 1 .
Step 3: However, the inner parenthesis didn’t simply say,
“ .” Replace with original values. = 4(1 − 3 ) . Solution: 1 ()
= − 3 + 1 2 − 3 + 1
Step 4: Differentiate inner value. The derivative of 1 − 3 is
“−3.” = 5 − 3 + 1 3 − 3
= 15 − 3 + 1 − 1
Step 5: Combine. = −3 4(1 − 3 ) .

Step 6: Simplify: = −12(1 − 3 ) .

Example 11:
Find the derivative of =   2 + 1.

Solution 1: Solution 2:
=
= = 2 + 1 = 2 + 1 1 ()
2 2 + 1

= 2 + 1 =
= 2 = 2 = 2 2 + 1 2 = 4 2 + 1

= 2 = 2

= 4(2 + 1) + 2 + 1 2
= × = 2 2 = 4(2 + 1)
= 2 2 + 1 2 + 2 + 1

= 4(2 + 1) + 2 + 1 2 = 2 2 + 1 4 + 1

= 2 2 + 1 2 + 2 + 1

= 2 2 + 1 4 + 1

Example 12: .
Find the derivative of =

Solution 1: Solution 2:
= 1 − 2
= (1 − 2 ) = 3 + 1 1 () = 3 + 1
= 2 1 − 2 2 1 − 2
= 1 − 2 = = 3 −2 = −4 1 − 2 = 3


= −2 = 2

′ = 3 + 1 −4(1 − 2 ) − (1 − 2 ) 3
= × = 2 −2 = −4 1 − 2
(3 + 1)

= 3 + 1 −4(1 − 2 ) − (1 − 2 ) 3 (1 − 2 ) −12 − 4 − 3 + 6
= (3 + 1)
(3 + 1)

(1 − 2 ) −12 − 4 − 3 + 6 (2 − 1)(6 + 7)
= (3 + 1) = (3 + 1)

(2 − 1)(6 + 7)
= (3 + 1)

Solution:

Use the quotient rule:

2 + 1 2 − 2 2 2
= = (2 + 1)
(2 + 1)

Example 13: Rewrite the 1st derivative before finding the 2nd derivative:

Find the first and second = 2(2 + 1)
derivatives of = . Hence,
Use chain rule (general power rule):
show that + = 0.
8
= −4 2 + 1 2 = − (2 + 1)

Substitute the 1st and 2nd derivatives into the left side of the equation:

8 2
+ = − (2 + 1) + 2 + 1

8 8
= − 2 + 1 + 2 + 1
= 0 shown .