Hey there! Sign in to join this conversationNew here? Join for free

differentiation Watch

Announcements
    • Thread Starter
    Offline

    2
    ReputationRep:
    given that y = x^2/ ax+b (a, b doesnt equal 0) show tat d^2y/dx^2 = 2b^2/(ax+b)^3

    dy/dx= ax^2 + 2bx / (ax+b)^2

    where do i go from here
    • Community Assistant
    • Welcome Squad
    Offline

    19
    ReputationRep:
    (Original post by Custardcream000)
    given that y = x^2/ ax+b (a, b doesnt equal 0) show tat d^2y/dx^2 = 2b^2/(ax+b)^3

    dy/dx= ax^2 + 2bx / (ax+b)^2

    where do i go from here
    Assuming it's correct (I can't be bothered to check) then just differentiate it again.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    Assuming it's correct (I can't be bothered to check) then just differentiate it again.
    so d^2y/dx^2 = [(ax+b)^2(2ax+2b) - (ax^2 +2bx)(2a(ax+b))]/ (ax+b)^4
    • Community Assistant
    • Welcome Squad
    Offline

    19
    ReputationRep:
    (Original post by Custardcream000)
    so d^2y/dx^2 = (ax+b)^2(2ax+2b) - (ax^2 +2bx)(2a)/ (ax+b)^4
    Nope. Check it again and post your working if you get stuck.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    Nope. Check it again and post your working if you get stuck.
    d^2y/dx^2 = (ax+b)^2(2ax+2b) - (ax^2 +2bx)(2a(ax+b))/ (ax+b)^4
    • Community Assistant
    • Welcome Squad
    Offline

    19
    ReputationRep:
    (Original post by Custardcream000)
    d^2y/dx^2 = (ax+b)^2(2ax+2b) - (ax^2 +2bx)(2a(ax+b))/ (ax+b)^4
    No...

    You shouldn't even get a power of 4.
    Offline

    3
    ReputationRep:
    (Original post by Custardcream000)
    d^2y/dx^2 = (ax+b)^2(2ax+2b) - (ax^2 +2bx)(2a(ax+b))/ (ax+b)^4
    It might be easier to follow if you work it out on paper and then post a picture of your working.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    No...

    You shouldn't even get a power of 4.
    i'm using the (vu' - uv')/v^2 method though
    Offline

    20
    ReputationRep:
    please put brackets around the top of the fraction.
    • Community Assistant
    • Welcome Squad
    Offline

    19
    ReputationRep:
    (Original post by Custardcream000)
    i'm using the vu' - uv'/v^2 method though
    In which case one power should cancel and it's hard to see where without your working.

    u=ax^2+2bx \Rightarrow u'=2ax+2b

    v=(ax+b)^2 \Rightarrow v'=2a(ax+b)

    \displaystyle \frac{vu'-uv'}{v^2}=\frac{(2ax+2b)(ax+b)^2-(ax^2+bx)[2a(ax+b)]}{(ax+b)^4}

    You can factor out one ax+b from the numerator straight away and cancel a power. The rest is just tidying up via expansion and cancellation.

    Please use latex or square brackets next time when writing out your fractions, or show your working, I got confused when looking at them.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    In which case one power should cancel and it's hard to see where without your working.

    u=ax^2+2bx \Rightarrow u'=2ax+2b

    v=(ax+b)^2 \Rightarrow v'=2a(ax+b)

    \displaystyle \frac{vu'-uv'}{v^2}=\frac{(2ax+2b)(ax+b)^2-(ax^2+bx)[2a(ax+b)]}{(ax+b)^4}

    You can factor out one ax+b from the numerator straight away and cancel a power. The rest is just tidying up via expansion and cancellation.

    Please use latex or square brackets next time when writing out your fractions, or show your working, I got confused when looking at them.
    thanks so i get

    {2a^3x^2 + 4a^2bx^2 + 6ab^2x +2b^3 - 2a^2x^2 + 2abx} / (ax+b)^3
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Should Spain allow Catalonia to declare independence?
    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
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

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