C4 Chapter 8: Differentiation
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
www.drfrostmaths.com
Last modified: 30th August 2015
Recap
Previously, in C1, you have only been able to differentiate
expressions of the form 𝑎𝑥 𝑏 + ⋯. Your differentiation skills will be
significantly expanded by the end of the chapter!
𝑑
3
𝟑 −𝟏
3 𝑥+
= 𝒙 𝟐 −? 𝟑𝒙−𝟐
𝑑𝑥
𝑥
𝟐
𝟏
𝑑 3 + 4𝑥 2
𝟑 −𝟑
= − 𝒙 𝟐? + 𝟔𝒙𝟐
𝑑𝑥
𝟐
𝑥
Rule #1: The Chain Rule
The Chain Rule allows you to differentiate when you have a composite function, i.e. a
function of a function.
𝑦 = 3𝑥 4 + 𝑥
5
! Chain Rule: If 𝑏𝑙𝑎 is the inner function, times the outer
function differentiated (w.r.t 𝑏𝑙𝑎) by 𝑏𝑙𝑎 differentiated.
𝑑𝑦
More formally if 𝑦 = 𝑓 𝑔 𝑥 then 𝑑𝑥 = 𝑓 ′ 𝑔 𝑥 ?𝑔′ 𝑥
𝑑𝑦
𝑑𝑦
𝑑𝑢
Or 𝑑𝑥 = 𝑑𝑢 × 𝑑𝑥 (where 𝑢 is the ‘bla’)
If we call this inner function ‘bla’, the outer function is 𝑏𝑙𝑎5 .
Differentiate this outer function (in terms of 𝑏𝑙𝑎), we have
5 𝑏𝑙𝑎4 = 5 3𝑥 4 + 𝑥 5
Now differentiating the inner function 𝑏𝑙𝑎, we have:
12𝑥 3 + 1
Multiplying these together gives the overall result:
𝑑𝑦
= 5 3𝑥 4 + 𝑥 4 12𝑥 3 + 1
𝑑𝑥
More Examples
Differentiate 𝑏𝑙𝑎4 then
times by 𝑏𝑙𝑎 differentiated.
𝑦=
𝑥3
𝑦= 𝑥
+ 𝑥
10
+𝑥
1
𝑦=
𝑥+1
𝑦=
1
3𝑥 + 1
4
11 12
𝑑𝑦
= 4 𝑥3 + 𝑥
𝑑𝑥
𝑑𝑦
= 12 𝑥 10 + 𝑥 11
𝑑𝑥
𝑑𝑦
= −1 𝑥 + 1
𝑑𝑥
𝑑𝑦
1
= − 3𝑥 + 1
𝑑𝑥
2
−2
3
−2
3𝑥 2
?
1 −1
+ 𝑥 2
2
11? 10𝑥 9
+ 11𝑥 10
1
1 ?= −
𝑥+1
3? = −
2
3
2 3𝑥 + 1
3
2
Test Your Understanding
So do you get it?
Q
Determine the gradient of the function below when 𝑥 = 3:
1
𝑦= 3
𝑥 + 4𝑥 2
𝑑𝑦
= −3 𝑥 3 + 4𝑥 2
𝑑𝑥
When 𝑥 = 3, 𝑚 = −
−4
17
1750329
3
2 + 8𝑥
3
3𝑥
3𝑥 2 + 8𝑥 = − 3
𝑥 + 4𝑥 2 4
?
= −9.71 × 10−6
Rule #2: 𝑑𝑥/𝑑𝑦
You’ve been used to having 𝑦 in terms of 𝑥 and finding
But we can find
𝑑𝑥
𝑑𝑦
𝑑𝑦
𝑑𝑥
even when 𝑥 is in terms of 𝑦.
!
𝑑𝑦
𝑑𝑥
=
1
Click for
Proof
𝑑𝑥
𝑑𝑦
𝑑𝑦
Find 𝑑𝑥 when 𝑥 = 2𝑦 2 + 𝑦
𝒅𝒙
= 𝟒𝒚 + 𝟏
𝒅𝒚
?𝟏
𝒅𝒚
=
𝒅𝒙 𝟒𝒚 + 𝟏
Find the gradient of 𝑥 = 1 + 2𝑦 3 when 𝑦 = 1
𝒅𝒙
𝒅𝒚
𝟏
𝟐
= 𝟔 𝟏 + 𝟐𝒚
→
=
𝒅𝒚
𝒅𝒙 𝟔 𝟏 + 𝟐𝒚
𝟏
𝟏
?
𝒎=
=
𝟔 𝟏 + 𝟐 𝟐 𝟓𝟒
𝟐
Exercise 8A
1
Differentiate:
4
→
a 1 + 2𝑥
3 + 4𝑥
1
e 3 + 2𝑥
c
h 3 8−𝑥
1
2
→
𝟖 𝟏 +?𝟐𝒙
𝟏
−
?
𝟐 𝟑 + 𝟒𝒙 𝟐
→ −
−6
𝟑
𝟐
? 𝟐𝒙
𝟑+
→ 𝟏𝟖 𝟖 −?𝒙
3
𝟐
−𝟕
𝑑𝑦
3
Given that 𝑦 = 5 − 2𝑥
−𝟓𝟒 ?
5
Find the value of 𝑑𝑥 at the point 2 2 , 4 on the curve with
𝑑𝑦
1
2
equation 𝑦 + 𝑦
𝟏
𝟓
?
𝟑
1
−2
=𝑥
find the value of 𝑑𝑥 at 1,27
1
Just before the next rule…
Sometimes in C3 you get expressions which you may be asked to tidy up by factorising.
Simplify 4𝑥 2 2𝑥 + 1
50
+ 2𝑥 2𝑥 + 1
51
→ 𝟐𝒙 𝟐𝒙 + 𝟏 𝟓𝟎 𝟐𝒙 + 𝟐𝒙 + 𝟏
?
= 𝒙 𝟐𝒙 + 𝟏 𝟓𝟎 𝟒𝒙 + 𝟏
Simplify 1 + 𝑥
1
2
+ 1+𝑥
1
2
−
= 𝟏+𝒙
= 𝟏+𝒙
=
Test Your Understanding:
1
−3
3𝑥 2𝑥 + 1
+ 2𝑥 + 1
2
Bro Tip: We factor
out the lowest power
of each term.
𝟏
−
𝟐
𝟏
−𝟐
𝟏+𝒙 +𝟏
𝟐+𝒙
?
𝟐+𝒙
𝟏+𝒙
−2
𝟏
→
𝟐𝒙 + 𝟏
𝟐
−𝟑
?+𝟏
𝟔𝒙 + 𝟐𝒙
𝟖𝒙 + 𝟏
=
𝟐 𝟐𝒙 + 𝟏
𝟑
Rule #3: Product Rule
! Product Rule: If 𝒚 = 𝒖𝒗 then
𝒅𝒚
𝒅𝒙
=𝒖
𝒅𝒗
𝒅𝒙
+𝒗
𝒅𝒖
𝒅𝒙
Click for Proof
In words: To differentiate the product of two functions, differentiate the first function
and leave the second alone, then vice versa, and add the two products together.
100
𝑑𝑦
E1
If 𝑦 = 𝑥 2 𝑥 + 1
E2
Given that 𝑓 𝑥 = 𝑥 2 3𝑥 − 1 determine 𝑓 ′ 𝑥 , simplifying your answer as much as
possible.
𝟏
𝟏
𝟏
−𝟐
′
𝟐
𝒇 𝒙 =𝒙
𝟑𝒙 − 𝟏
𝟑 + 𝟐𝒙 𝟑𝒙 − 𝟏 𝟐
𝟐
𝟏
𝟏
Be careful here.
−𝟐
= 𝒙 𝟑𝒙 − 𝟏
𝟑𝒙 + 𝟒?𝟑𝒙 − 𝟏
𝟐
𝟏
𝟏
−𝟐
= 𝒙 𝟑𝒙 − 𝟏 (𝟏𝟓𝒙 − 𝟒)
𝟐
, determine , simplifying your answer.
𝑑𝑥
𝒅𝒚
= 𝒙𝟐 𝟏𝟎𝟎 𝒙 + 𝟏 𝟗𝟗 + 𝟐𝒙 𝒙 + 𝟏 𝟏𝟎𝟎
𝒅𝒙
?
𝟗𝟗
𝟐
= 𝒙+𝟏
𝟏𝟎𝟎𝒙 + 𝟐𝒙 𝒙 + 𝟏 = 𝟐𝒙 𝒙 + 𝟏 𝟗𝟗 𝟓𝟏𝒙 + 𝟏
Test Your Understanding
Differentiate the following:
1
𝑦 =𝑥 𝑥+1
𝟏
𝒅𝒚
𝟏
−
=𝒙
𝒙+𝟏 𝟐 +𝟏 𝒙+𝟏
𝒅𝒙
𝟐
𝟏
𝟏
−
= 𝒙+𝟏 𝟐 𝒙+
?𝟐 𝒙 + 𝟏
𝟐
𝟑𝒙 + 𝟐
=
𝟐 𝒙+𝟏
2
𝑦 = 2𝑥 + 1 3 3𝑥 + 1 2
𝒅𝒚
= 𝟑 𝟐𝒙 + 𝟏 𝟐 𝟐 𝟑𝒙 + 𝟏 𝟐 + 𝟐 𝟑𝒙 + 𝟏 𝟑 𝟐𝒙 + 𝟏
𝒅𝒙
= 𝟔 𝟐𝒙 + 𝟏 𝟐 𝟑𝒙 + 𝟏 𝟐 + 𝟐𝒙 + 𝟏 𝟑 𝟑𝒙 + 𝟏
? 𝟏 + 𝟐𝒙 + 𝟏
= 𝟔 𝟐𝒙 + 𝟏 𝟐 𝟑𝒙 + 𝟏 𝟑𝒙 +
= 𝟔 𝟐𝒙 + 𝟏 𝟐 𝟑𝒙 + 𝟏 𝟓𝒙 + 𝟐
𝟏
𝟐
𝟑
Exercise 8B
5
1a
𝑥 1 + 3𝑥
1c
𝑥 3 2𝑥 + 6
1d
3𝑥 2 5𝑥 − 1
2b
4
−1
Find the value of
13
+ 𝟏 + 𝟑𝒙 ?𝟓 = 𝟏 + 𝟑𝒙
→ 𝟏𝟓𝒙 𝟏 + 𝟑𝒙
𝟒
→ 𝟐𝒙𝟐 𝟐𝒙 + 𝟔
𝟕𝒙 + 𝟗
𝟑
→ 𝟑𝒙 𝟓𝒙 − 𝟐 𝟓𝒙 − 𝟏
𝑑𝑦
𝑑𝑥
−𝟐
𝟒
𝟏𝟖𝒙 + 𝟏
?
?
1
2
at the point 4,36 on the curve with equation 𝑦 = 3𝑥 2𝑥 + 1 .
?
3
Find the points where the gradient is zero on the curve with equation
𝑦 = 𝑥 − 2 2 2𝑥 + 3 .
Examiner’s Report: “Perhaps the most challenging question
𝟏
𝟏𝟗
(𝟐, 𝟎) and −?𝟑 , 𝟏𝟐 𝟐𝟕
June 2013 Q5(c)
𝑑𝑦
=
𝑑𝑥
Find
1
6𝑥 𝑥 − 1
𝑑2 𝑦
,
𝑑𝑥 2
1
2
simplifying your answer.
in the paper”. And this is on a paper described as having “a
level of demand higher than any previous C3 paper”.
So it’s damn hard.
𝟏
𝒅𝒚 𝟏 −𝟏
= 𝒙 𝒙 − 𝟏 −𝟐
𝒅𝒙 𝟔
𝟏
𝒅𝟐 𝒚
𝟏 −𝟐
𝟏 −𝟏
−
𝟐−
=
−
𝒙
𝒙
−
𝟏
𝒙 𝒙−𝟏
𝒅𝒙𝟐
𝟔
𝟏𝟐
𝟑
𝟏 −𝟐
−
=−
𝒙 𝒙−𝟏 𝟐 𝟐 𝒙−𝟏 +𝒙
𝟏𝟐
𝟐 − 𝟑𝒙
=
𝟑
𝟏𝟐𝒙𝟐 𝒙 − 𝟏 𝟐
?
−
𝟑
𝟐
The Quotient Rule
We can use the ‘product’ rule to differentiate a product 𝑦 = 𝑢𝑣
𝑢
How can we differentiate a ‘quotient’, i.e. 𝑦 = 𝑣
Using the Chain and Product rules:
𝑢
?
= 𝑢𝑣 −1
𝑣
𝑑𝑦
𝑑𝑣 𝑑𝑢 −1
= −𝑢𝑣 −2 ?
+
𝑣
𝑑𝑥
𝑑𝑥 𝑑𝑥
𝑑𝑢
𝑑𝑣
= 𝑣 −2 𝑣 ?− 𝑢
𝑑𝑥
𝑑𝑥
𝑦=
=
This is the Quotient Rule. It
looks like the Product Rule
except the + is now a −
and we divide by the
denominator squared.
𝑣
𝑑𝑢
𝑑𝑣
−𝑢
𝑑𝑥? 𝑑𝑥
𝑣2
Bro Tip: How do you
𝑑𝑢
remember whether 𝑣 or
𝑑𝑣
𝑢 𝑑𝑥
𝑑𝑥
the
comes first?
I remember “VDU (Visual
Display Unit) first”!
Examples
𝑑𝑢 𝑑𝑣
𝑥+1
𝑦= 2
3𝑥 + 𝑥
𝑥
𝑦=
𝑥+1
2𝑥 + 1
𝑦= 2
𝑥 −1
𝑢
𝑣
Bro Tip: You may wish to write out 𝑢, 𝑣, , before applying the
𝑑𝑥 𝑑𝑥
Quotient Rule, but I would only bother if you’re liable to silly mistakes.
𝑑𝑦
3𝑥 2 + 𝑥 − 𝑥 + 1 6𝑥 + 1
=
?
𝑑𝑥
3𝑥 2 + 𝑥 2
1 −12
𝑑𝑦 2 𝑥 𝑥 +?1 − 𝑥
=
𝑑𝑥
𝑥+1 2
1−𝑥
=
?2
2 𝑥 𝑥+1
𝑑𝑦
=
𝑑𝑥
𝑥2
1
−2
− 1 2𝑥 + 1
− 2𝑥 2𝑥 + 1
𝑥 2 −?1 2
We’ll practice this later once we’ve learnt how to differentiate 𝑠𝑖𝑛, 𝑒 𝑥 , etc…
1
2
Differentiating 𝑒 𝑥 and ln 𝑥
𝑑 𝑥
𝑒 = 𝑒 𝑥?
𝑑𝑥
𝑑
1
ln 𝑥 = ?
𝑑𝑥
𝑥
Click for Proof
Click for Proof
This is probably one of the most
important results in calculus,
and is why 𝑒 comes up
everywhere!
You know how to find 𝑥 −2 𝑑𝑥 in
the usual way. But what if you tried
to do 𝑥 −1 𝑑𝑥 in the usual way?
Now our problem is solved!
Use the Chain Rule to differentiate the following:
“e to the bla” differentiated is just “e to the bla”.
𝑑 𝑥2
𝑥2
𝑒
= 2𝑥𝑒
𝑑𝑥
“ln of bla” differentiated is “1 over bla”
𝑑
ln 𝑥 2 + 1
𝑑𝑥
2𝑥
= 2
𝑥 +1
A Few More Examples
𝑦 = 𝑒 2𝑥
→
𝑦 = 𝑒 3𝑥
𝒅𝒚
= 𝟐𝒆𝟐𝒙
𝒅𝒙
2 +1
→
𝒅𝒚
?𝟐+𝟏
= 𝟔𝒙 𝒆𝟑𝒙
𝒅𝒙
𝒅𝒚 𝟓
=
𝒅𝒙 𝒙
?
𝑦 = 5 ln 𝑥
→
𝑒 𝑥+1
𝑦=
𝑥
𝒅𝒚 𝒆𝒙+𝟏 𝒙 − 𝟏
?
→
=
𝟐
𝒅𝒙
𝒙
𝑦 = ln
𝑥3
𝑦 = 2𝑥 +
𝑦 = 𝑒𝑒
𝑥
+ 𝑒𝑥
𝑒𝑥
ln 𝑥
Bro Tip: We can see in general
𝑑
that 𝑑𝑥 𝑒 𝑓 𝑥 = 𝑓 ′ 𝑥 𝑒 𝑓 𝑥 ,
i.e. we just multiply by the
differentiated power.
𝒅𝒚 𝟑𝒙𝟐 + 𝒆𝒙
?
→
= 𝟑
𝒙
𝒅𝒙
𝒙 +𝒆
𝒅𝒚
𝒆𝒙
𝟏 𝒙
𝒙
→
=𝟐+ +
? 𝒆 𝐥𝐧 𝒙 = 𝟐 + 𝒙 𝒆 𝟏?+ 𝒙 𝐥𝐧 𝒙
𝒅𝒙
𝒙
→
𝒅𝒚
𝒙
𝒙
= 𝒆𝒙 𝒆𝒆 =?𝒆𝒙+𝒆
𝒅𝒙
Exercises
1
𝑦 = ln 2𝑥 + 1
2
𝑦=𝑒
3
𝑦 = ln 𝑥
4
𝑒𝑥
𝑦= 2
𝑥 +1
5
𝑦=
𝑒𝑥
6
𝑦 = ln 𝑒 𝑥 + 1
7
𝑦=
𝑥+1 5
10
ln 𝑥
1
2
𝑥 𝑒𝑥
𝟐
?
𝟐𝒙 + 𝟏
→ 𝟓 𝒙 + 𝟏?𝟒 𝒆
𝟏𝟎 𝐥𝐧 𝒙 𝟗
?
→
𝒙
𝒙
𝒆 𝒙−𝟏 𝟐
→
?
𝒙𝟐 + 𝟏 𝟐
8
→
ln 𝑥
→
𝟐𝒙 ?
−𝟏
with equation 𝑦 =
.
𝑥
𝟏
𝒙
− 𝐥𝐧 𝒙 𝟏 − 𝐥𝐧 𝒙
𝒅𝒚
𝒙
=
=
𝒅𝒙
𝒙𝟐
𝒙𝟐
𝒙+𝟏 𝟓
𝟏 𝒙
→ 𝒆 𝟏+
? 𝒙 𝐥𝐧 𝒙
𝒙
𝒆𝒙
→ 𝒙
?
𝒆 +𝟏
𝟏
𝒆𝒙
Find the turning point of the curve
𝟏 − 𝐥𝐧 𝒙 = 𝟎
𝐥𝐧 𝒙 = 𝟏
𝒙=𝒆
𝟏
𝒚=
𝒆
?
9
Find the equation of the tangent to the
𝑒𝑥
curve with equation 𝑦 =
when
ln 𝑥
𝑥 = 𝑒.
𝟏 𝒙
𝟐
𝒙
𝒅𝒚 𝐥𝐧 𝒙 𝒆 − 𝒙 𝒆
𝒆𝒙 𝐥𝐧 𝒙 − 𝟏
=
=
𝒅𝒙
𝐥𝐧 𝒙 𝟐
𝒙 𝐥𝐧 𝒙 𝟐
𝒅𝒚
When 𝒙 = 𝒆, = 𝟎
𝒅𝒙
So tangent: 𝒚 = 𝒆𝒆
?
Differentiating Trig Functions
𝑑
sin 𝑥 = cos?𝑥
𝑑𝑥
Click for Proof
Bro Tip: The way I picture this in
my head is this:
𝑑
? 𝑥
cos 𝑥 = − sin
𝑑𝑥
sin 𝑥
cos 𝑥
where moving ‘down’ is
differentiating and moving up is
integrating. If it’s the ‘wrong’
function (differentiating cos or
integrating sin) the sign changes.
Examples
𝑑
cos 𝑥 2
𝑑𝑥
𝑑
sin 2𝑥 = 𝟐 𝐜𝐨𝐬?𝟐𝒙
𝑑𝑥
𝑑 2
𝑥 sin 𝑥 = 𝒙𝟐 𝐜𝐨𝐬 𝒙 +
?𝟐𝒙 𝐬𝐢𝐧 𝒙
𝑑𝑥
? 𝒙𝟐
= −𝟐𝒙 𝐬𝐢𝐧
𝑑 cos 𝑥
𝑒
= − 𝐬𝐢𝐧 𝒙? 𝒆𝐜𝐨𝐬 𝒙
𝑑𝑥
Edexcel C3 June 2013 (R) Q5a
Differentiate with respect to 𝑥:
cos 2𝑥
𝑥
𝑢 = cos 2𝑥
v= 𝑥
𝑑𝑢
𝑑𝑣 1 −1
= −2 sin
? 2𝑥 𝑑𝑥 = 2 𝑥 2
𝑑𝑥
1 −12
𝑑𝑦 −2 𝑥 sin 2𝑥 − 2 𝑥 cos 2𝑥
1 −3
=
= − 𝑥 2 4𝑥 sin 2𝑥 + cos 2𝑥
𝑑𝑥
𝑥
2
Differentiating Other Trig Functions
These can all be derived, but you really need to memorise them – they’re NOT in your
formula booklet.
𝑑
PROOF ?
tan 𝑥 = sec?2 𝑥
𝑑𝑥
𝑑
cosec 𝑥 = − c𝑜𝑠𝑒𝑐?𝑥 cot 𝑥
𝑑𝑥
sin 𝑥
Using Quotient Rule on tan 𝑥 =
:
cos 𝑥
𝑑 sin 𝑥
cos2 𝑥 + sin2 𝑥
1
=
=
= sec 2 𝑥
2
2
𝑑𝑥 cos 𝑥
cos 𝑥
cos 𝑥
𝑑
sec 𝑥 = sec 𝑥 ?tan 𝑥
𝑑𝑥
𝑑
? 2𝑥
cot 𝑥 = −𝑐𝑜𝑠𝑒𝑐
𝑑𝑥
Quickfire…
𝑑
sec 𝑥 = sec 𝑥 ?tan 𝑥
𝑑𝑥
𝑑
? 𝑥
cos 𝑥 = − sin
𝑑𝑥
Quickfire…
𝑑
tan 𝑥 = sec?2 𝑥
𝑑𝑥
𝑑
? 2𝑥
cot 𝑥 = −𝑐𝑜𝑠𝑒𝑐
𝑑𝑥
Quickfire…
𝑑
cosec 𝑥 = − c𝑜𝑠𝑒𝑐?𝑥 cot 𝑥
𝑑𝑥
𝑑
sin 𝑥 = cos?𝑥
𝑑𝑥
Quickfire…
Click to Repeat >
𝑑
sec 𝑥 = sec 𝑥 ?tan 𝑥
𝑑𝑥
𝑑
? 𝑥
cos 𝑥 = − sin
𝑑𝑥
Click to Escape >
Examples
Differentiate the following:
cosec 2𝑥
𝑦=
𝑥2
𝑦 = sec 3 𝑥
𝑑𝑦 𝑥 2 −2 csc 2𝑥 cot 2𝑥 − csc 2𝑥 𝑥 2
=
4
𝑑𝑥
𝑥
?
−2 csc 2𝑥 𝑥 cot 2𝑥 + 1
=
𝑥3
𝑦 = sec 3 𝑥 = sec 𝑥 3 Using chain rule
𝑑𝑦
= 3 sec 2 𝑥 ?sec 𝑥 tan 𝑥
𝑑𝑥
= 3 sec 3 𝑥 tan 𝑥
Exercises
Edexcel Jan 2011 Q7
Exercise 8I
1
June 2013
3+sin 2𝑥
The curve 𝐶 has equation 𝑦 = 2+cos 2𝑥.
𝟐
𝟒𝒙
a 𝑦 = cot 4𝑥 → −𝟒 𝐜𝐬𝐜?
c 𝑦 = cosec 4𝑥 → −𝟒 𝐜𝐬𝐜 𝟒𝒙
? 𝐜𝐨𝐭 𝟒𝒙
d 𝑦 = sec 2 3𝑥 → 𝟔 𝐬𝐞𝐜 𝟐 𝟑𝒙
? 𝐭𝐚𝐧 𝟑𝒙
sec 2 𝑥
𝐬𝐞𝐜 𝟐 𝒙 𝟐𝒙 𝐭𝐚𝐧 𝒙 − 𝟏
→
f 𝑦= 𝑥
𝒙𝟐
3
𝟑
g 𝑦 = cosec 2𝑥 → −𝟔 𝐜𝐬𝐜 𝟑𝒙 𝐜𝐨𝐭 𝟐𝒙
𝑦 = cot 2 2𝑥 − 1
?
?
h
?
→ −𝟒 𝒄𝒐𝒕 𝟐𝒙 − 𝟏 𝒄𝒔𝒄𝟐 𝟐𝒙 − 𝟏
Exercise 8J
3
b 𝑦 = 𝑥 2 sec 3𝑥
𝑑𝑦
Show that 𝑑𝑥 =
6 sin 2𝑥+4 cos 2𝑥+2
2+cos 2𝑥 2
𝒅𝒚 (𝟐 + 𝐜𝐨𝐬 𝟐𝒙) 𝟐 𝐜𝐨𝐬 𝟐𝒙 − 𝟑 + 𝐬𝐢𝐧 𝟐𝒙 −𝟐 𝐬𝐢𝐧 𝟐𝒙
=
𝒅𝒙
𝟐 + 𝐜𝐨𝐬 𝟐𝒙 𝟐
?
=⋯
Find an equation of the tangent to 𝐶 at the point on 𝐶
𝜋
where 𝑥 = . Write your answer in the form 𝑦 =
2
𝑎𝑥 + 𝑏 where 𝑎 and 𝑏 are constants.
𝟔 𝐬𝐢𝐧 𝝅 + 𝟒 𝐜𝐨𝐬 𝝅 + 𝟐 𝟎 − 𝟒 + 𝟐
𝒎=
=
= −𝟐
𝟐 + 𝐜𝐨𝐬 𝝅 𝟐
𝟐−𝟏 𝟐
𝟑 + 𝐬𝐢𝐧 𝝅
𝒚=
=𝟑
𝟐 + 𝐜𝐨𝐬 𝝅
𝝅
𝒚 − 𝟑 = −𝟐 𝒙 −
𝟐
𝒚 = −𝟐𝒙 + 𝟑 + 𝝅
?
?
→ 𝟐𝒙 𝐬𝐞𝐜 𝟑𝒙 + 𝟑𝒙𝟐 𝐬𝐞𝐜 𝟑𝒙 𝐭𝐚𝐧 𝟑𝒙
d 𝑦 = sin3 𝑥 cos 𝑥
→ 𝟑 𝐬𝐢𝐧𝟐 𝒙 𝒄𝒐𝒔𝟐 𝒙 − 𝒔𝒊𝒏𝟒 𝒙
1 + sin 𝑥
𝟏 + 𝐬𝐢𝐧 𝒙
𝑦
=
→
f
cos 𝑥
𝐜𝐨𝐬 𝟐 𝒙
ln 𝑥
𝐭𝐚𝐧 𝒙 − 𝒙 𝐬𝐞𝐜 𝟐 𝒙 𝐥𝐧 𝒙
→
k 𝑦=
tan 𝑥
𝒙 𝐭𝐚𝐧𝟐 𝒙
𝑒 sin 𝑥
𝒆𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬𝟐 𝒙 + 𝐬𝐢𝐧 𝒙
l 𝑦=
→
cos 𝑥
𝐜𝐨𝐬 𝟐 𝒙
?
?
?
?
Edexcel Jan 2012 Q4
The point 𝑃 is the point on the curve 𝑥 = 2 tan 𝑦 +
𝜋
𝜋
12
with 𝑦-coordinate . Find an equation of the normal to
4
the curve at 𝑃. (7 marks)
𝝅
Bro Tip: If you’re not asked to
𝒚 − = −𝟖 𝒙 − 𝟐 𝟑
simplify, then don’t bother!
𝟒
?
APPENDIX
Proofs of derivatives
Proof that ln 𝑥 → 1/𝑥
Note first that 𝒆 = 𝐥𝐢𝐦 𝟏 +
𝒏→∞
𝟏 𝒏
.
𝒏
1000
1
1000
For example, try entering 1 +
onto your calculator! You may wish to look into
Bernoulli’s Compound Interest Problem in your own time.
This is how we found the
gradient function ‘by first
principles’ in C1.
If 𝑓 𝑥 = ln 𝑥
𝑓 𝑥+ℎ −𝑓 𝑥
ln 𝑥 + ℎ − ln 𝑥
= lim
ℎ→0
ℎ→0
ℎ
ℎ
𝑓 ′ 𝑥 = lim
1
𝑥+ℎ
= lim ln
ℎ→0 ℎ
𝑥
= lim ln 1 +
ℎ→0
Letting 𝑛 =
ℎ
𝑥
𝑥+ℎ
= lim ln
ℎ→0
𝑥
1
ℎ
By using laws of logs.
This is a sensible substitution because it gives us something
resembling the definition of 𝑒. Note that as ℎ → 0, 𝑛 → ∞.
𝑥
:
ℎ
1
= lim ln 1 +
𝑛→∞
𝑛
1
ℎ
1
𝑛
𝑥
1
1
= lim ln 1 +
𝑛→∞ 𝑥
𝑛
= lim ln
𝑛→∞
𝑛
1
1+
𝑛
1
1
= 𝑙𝑛 lim 1 +
𝑥 𝑛→∞
𝑛
𝑛
𝑛
=
1
𝑥
The power can be moved back to the front by
𝑥
law of logs, and outside the limit since 𝑛 is the
limiting variable. It doesn’t matter whether we
find the limit before or after the log.
1
1
ln 𝑒 =
𝑥
𝑥
□
Proof that 𝑒 𝑥 → 𝑒 𝑥
𝑑
1
We will assume that we have already proven that 𝑑𝑥 ln 𝑥 = 𝑥 (see previous slide).
𝑦 = 𝑒𝑥
𝑥 = ln 𝑦
𝑑𝑥 1
=
𝑑𝑦 𝑦
𝑑𝑦
= 𝑦 = 𝑒𝑥 □
𝑑𝑥
Proof that sin 𝑥 → cos 𝑥
You need to know that for small values of 𝑥, sin 𝑥 ~𝑥 (i.e. are equal in
the limit). We can see this by comparing the areas of the sector and
triangle, which becomes increasingly similar as 𝑥 becomes small.
Similarly cos 𝑥 ~1 as 𝑥 becomes small.
𝑦 = sin 𝑥
𝑑𝑦
sin 𝑥 + ℎ − sin 𝑥
= lim
𝑑𝑥 ℎ→0
ℎ
sin 𝑥 cos ℎ + sin ℎ cos 𝑥 − sin 𝑥
= lim
ℎ→0
ℎ
sin 𝑥 + ℎ cos 𝑥 − sin 𝑥
= lim
ℎ→0
ℎ
ℎ 𝑐𝑜𝑠 𝑥
= lim
= lim cos 𝑥
ℎ→0
ℎ→0
ℎ
= cos 𝑥
□
Ridiculously Important Note: Since sin 𝑥 ~𝑥 is only valid when 𝑥 is
in radians (since we used the sector area formula for radians), then
𝑑
sin 𝑥 = cos 𝑥 is only applicable in radians!
𝑑𝑥
For this reason, you can assume we’re also in radians for calculus.
1
𝑥
1
Proof that cos 𝑥 → − sin 𝑥
𝑑
We will assume we have already proven that 𝑑𝑥 sin 𝑥 = cos 𝑥
Recall from C2 that cos 𝑥 = sin
𝑑
𝑑
Thus 𝑑𝑥 cos 𝑥 = 𝑑𝑥 sin
𝜋
2
−𝑥
𝜋
2
−𝑥
= − cos
𝜋
2
− 𝑥 = − sin 𝑥 □
by the Chain Rule.