Question

[A-level C3 Calculus] differentiate y=(6x-3)(2x-1)^5
Please Help!

Answer 1

You can take the natural log of both sides, and take the derivative that way.

Answer 2

okay how would i do that? the text book only mentions the quotient, chain and product rules though..?

Answer 3

can you apply quotient rule ?

Answer 4

are you sure this is calculus 3?

Answer 5

i'm not sure, could you help? and yeah its Core 3 AQA a level maths

Answer 6

what is quotient?

Answer 7

the quotient rule?

Answer 8

well I asked if you can apply quotient rule and you said you don't know
so I asked what is quotient

Answer 9

im really confused... so should i apply the quotient rule?

Answer 10

sorry im not making this easy for myself

Answer 11

\[\frac{p}{q} = pq^{-1}\]
this is quotient, correct?
do you see this format for you to apply quotient rule? no

Answer 12

the quotient rule: (g(x)f'(x) + g'(x)f(x) ) /g ^2

Answer 13

so that leaves us chain and product rule
let us start with product
\[a*b \]
is product, so we can think a = (6x-3) and b = (2x-1)^5

Answer 14

okay

Answer 15

so what about chain rule?

Answer 16

da/dx =-3 (sorry its (6-3x) in the first bracket

Answer 17

it doesn't matter we can deal with specifics later, we are just trying to understand what rules we can possibly apply by looking at the formats

Answer 18

oh okay yeah

Answer 19

the chain rule: dy/dx=du/dx * dy/du

Answer 20

(which can be extended by adding more functions)

Answer 21

so we can basically apply both product and chain rule
but we need to apply chain rule first

Answer 22

right.. soo..

Answer 23

which one would it be?

Answer 24

do you know power rule?

Answer 25

no we havent covered it but i watched a university lecture on this topic and it said d/dx (f(x))^n = nf^(n-1) df/dx

Answer 26

if that makes any sense?

Answer 27

yes

Answer 28

so the function = the differentiation * the integration or something?

Answer 29

integration? no

Answer 30

okay so what does the power rule state then?

Answer 31

this is why I am asking if you know power rule first, it is important that you know this very well first

Answer 32

oh okay

Answer 33

so do you?

Answer 34

nope

Answer 35

\[f(x) = x^{n} \rightarrow f'(x) = nx^{n-1}\]

Answer 36

is that the power rule? oh okay yeah i understand that one

Answer 37

okay we had x^n
what if we had (2x+3)^n?
this is where chain rule comes in
first we try and do the power rule of everything in the parenthesis
n(2x+3)^(n-1)
then multiply it to the derivative of whatever is inside the parenthesis
n(2x+3)^(n-1) * (2)

Answer 38

okay yeah im with you on that

Answer 39

so going back with your problem, you can apply the chain rule first before you can apply the product rule

Answer 40

so applying this to the question we get: 10(2x+3)^4

Answer 41

y=(6x-3)(2x-1)^5
chain rule with (2x-1)^5
then whatever answer you get, apply a product rule with 6x-3

Answer 42

=f(x)g'(x) + g(x)f'(x)

Answer 43

something along those lines

Answer 44

=(6x-3)(10(2x+3)^4) + (2x-1)^5 (-3) ?

Answer 45

what is the result of your chain rule?

Answer 46

* my mistake it should be (2x-1) not (2x-3) in the first set of brackets

Answer 47

would i expand the brackets?

Answer 48

=(60x-30)(2x-1)^4 + 32x *-x *-3
=

Answer 49

you lost me

Answer 50

what is the result of your chain rule?

Answer 51

I've lost myself.. aha i dont know?

Answer 52

10(2x-1)^4

Answer 53

why 10?

Answer 54

you're only bringing down 5

Answer 55

5(2x-1)^4 * (2)

Answer 56

oh yeah, but you multiply by d(2x-1)/dx which equals 2, therefore 2*5=10??

Answer 57

oh okay

Answer 58

is that correct?

Answer 59

yes

Answer 60

okay good, so now i presume we need to use the product rule?

Answer 61

yes

Answer 62

which would be (6-3x)*(the answer from before) + (2-1)^5 * d(6-3x)/dx

Answer 63

just that alone?

Answer 64

(6x-3) * 10(2x-1)^4

Answer 65

like that?

Answer 66

so...
= 60x-30 * (2x-1)^4
?

Answer 67

but that would only be f(x)g'(x)
as oppposed to f(x)g'(x) + f'(x)g(x)
?

Answer 68

I think you're doing a lot of typos

Answer 69

yeah product rule is
u'v+uv'

Answer 70

yeah.. thats an easier format!

Answer 71

uv' = (6x-3) * 10(2x-1)^4

Answer 72

go on

Answer 73

and u'v = ....

Answer 74

u'v = (2x-1)^5 * -3 ?

Answer 75

u = 6x-3
v = (2x-1)^5

u'v + uv'
u'v = 6(2x-1)^5
uv' = (6x-3) 10(2x-1)^4

Answer 76

clear?

Answer 77

oh right yeah, sorry its (6-3x) in the brackets not (6x-3) but yes i understand the concept

Answer 78

laughing out loud okay

Answer 79

u = -3x+6 then?

Answer 80

.... therefore the grand answer is...

6(2x-1)^5 + (6x-3) 10(2x-1)^4

am i right?

Answer 81

sometimes, it might be less confusing if you did this
list your u and v
then list your u' and then v'

Answer 82

yeah, we did several questions and that was generally how i formatted them in my answers

Answer 83

then you follow all the formats of product, quotient and chain rules

Answer 84

aye i see

Answer 85

this kind of systematic approach will help you in calculus 2

Answer 86

just to complicate things.. the answer in the book says (2x-1)^4 (63-36x)

Answer 87

we are multiply buy a negative so there must be a negative somewhere

Answer 88

that's what i thought, there must be a negative

Answer 89

and a power of 5 due to it being uv'

Answer 90

but one last question, y=sin3xcos2x
a simple product rule question right?
but what is dsin3x/dx ?

Answer 91

is it 3 cos x ?

Answer 92

3 is constant
and derivative of sin is cos

Answer 93

its been a while since i've done maths before today!

Answer 94

oh is it just cosx

Answer 95

oh thanks for that!

Answer 96

you're welcome
I use product as u'v+uv'
and quotient as (u'v - uv')/v^2
see the format? it's simple and no alternating you follow a simple alphabetical order