Turn on thread page Beta
    • Thread Starter
    Offline

    5
    ReputationRep:
    Hey,
    This is like my first thread but got a quick question. It's on differentiation in C3.
    Using the product rule, differentiate x(1+3x)^5.
    Thanks for any help
    • Community Assistant
    Offline

    18
    ReputationRep:
    Community Assistant
    (Original post by rainbowc4k3)
    Hey,
    This is like my first thread but got a quick question. It's on differentiation in C3.
    Using the product rule, differentiate x(1+3x)^5.
    Thanks for any help
    What have you tried so far?
    Offline

    18
    ReputationRep:
    (Original post by rainbowc4k3)
    Hey,
    This is like my first thread but got a quick question. It's on differentiation in C3.
    Using the product rule, differentiate x(1+3x)^5.
    Thanks for any help
    You know that,

    (uv)' = u'v + uv'

    Can you make a start from there?
    • Thread Starter
    Offline

    5
    ReputationRep:
    (Original post by NotNotBatman)
    What have you tried so far?
    I've tried using the product rule and I've seen the markscheme:

    let u=x v= (1+ 3x)^5

    du/dx = 1 dv/dx= 5*3(1+3x)^4

    (product rule) ====> dy/dx = u(dv/dx)+v(du/dx)
    ---> dy/dx = x*15(1+3x)^4 + (1+3x)^5 *1
    I can do it up to here but then I get confused
    =((1+3x)^4)(15x+1+3x)
    =(1+3x)^4 *(1+18x)
    How did they move the 15x into the other bracket and where does the power of 5 go?
    • Thread Starter
    Offline

    5
    ReputationRep:
    (Original post by _gcx)
    You know that,

    (uv)' = u'v + uv'

    Can you make a start from there?
    I made a start, but I'm confused even though I've seen the MS
    Offline

    18
    ReputationRep:
    (Original post by rainbowc4k3)
    I've tried using the product rule and I've seen the markscheme:

    let u=x v= (1+ 3x)^5

    du/dx = 1 dv/dx= 5*3(1+3x)^4

    (product rule) ====> dy/dx = u(dv/dx)+v(du/dx)
    ---> dy/dx = x*15(1+3x)^4 + (1+3x)^5 *1
    I can do it up to here but then I get confused
    =((1+3x)^4)(15x+1+3x)
    =(1+3x)^4 *(1+18x)
    How did they move the 15x into the other bracket and where does the power of 5 go?
    Expanding the brackets again,

    (1+3x)^4(15x+(3x+1)) = 15x(1+3x)^4 + (3x+1)(3x+1)^4 = 15x(1+3x)^4 + (3x+1)^5

    Is it clearer now?
    • Community Assistant
    Offline

    18
    ReputationRep:
    Community Assistant
    (Original post by rainbowc4k3)
    I've tried using the product rule and I've seen the markscheme:

    let u=x v= (1+ 3x)^5

    du/dx = 1 dv/dx= 5*3(1+3x)^4

    (product rule) ====> dy/dx = u(dv/dx)+v(du/dx)
    ---> dy/dx = x*15(1+3x)^4 + (1+3x)^5 *1
    I can do it up to here but then I get confused
    =((1+3x)^4)(15x+1+3x)
    =(1+3x)^4 *(1+18x)
    How did they move the 15x into the other bracket and where does the power of 5 go?
    Factorise out (1+3x)^4.

     \frac{dy}{dx} = 15x(1+3x)^4 + (1+3x)^5  = 15x(1+3x)^4 + (1+3x)^4(1+3x) because of the power rule.
    Now factorise (1+3x)^4 and consider what each term is multiplied by.

    or it might be easier to see if we let (1+3x)^4 = a, then dy/dx = 15xa^4+a^5
    Could you factorise that and then substitute a in terms of (1+3x)^4 at the end?
    • Thread Starter
    Offline

    5
    ReputationRep:
    Oh wowwwwww...
    I finally get it, thank you guys so much for the help it's literally so easy now that I see it.
    Thanks guys much appreciated.
    Offline

    4
    ReputationRep:
    y = f(x)g(x), then [f(x)g(x)]’ = f’(x) g(x)+ f(x) g’(x)
    for x(1+3x)^5

    take f(x) as x

    take g(x) as (1+3x)^5

    then

    f'(x) = 1
    g'(x) = 15x(1+3x)^4

    then just substitute into rule.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 8, 2017

1,337

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.