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

c3 differentiation help needed - quotient rule Watch

    • Thread Starter
    Offline

    0
    ReputationRep:
    quotient rule

    so here's the question..
    y = x^2/ (1+x)^1/2

    and the answer is 4x-3x^2 / 2(1+x)^3/2

    now i got the numerator but halfway through the working, the solutions said that the numerator and denominator had to be multiplied by 2(1+x)^1/2

    and so resulting in with the overall denominator of 2(1+x)^3/2


    can someone please explain to me why it had to be multiplied by 2(1+x)^1/2? thank you
    Offline

    17
    ReputationRep:
    y=\frac{x^2}{(1+x)^{\frac{1}{2}}  }

    \frac{dy}{dx} = \frac{[(1+x)^{\frac{1}{2}} \times \frac{d}{dx}x^2] - [x^2 \times \frac{d}{dx} (1+x)^{\frac{1}{2}}]}{((1+x)^{\frac{1}{2}})^2}

    \frac{dy}{dx} = \frac{[2x(1+x)^{\frac{1}{2}}] - [x^2 \times \frac{1}{2}(1+x)^{-\frac{1}{2}}]}{1+x}

    You want to get rid of the (1+x)^{-\frac{1}{2}} on the numerator, which you can do by multiplying both the numerator and denominator by (1+x)^{\frac{1}{2}}. This is because

    x^{-a} \times x^{a} = x^{-a+a} = x^0 = 1
    Offline

    15
    ReputationRep:
    Well, the quotient rule is basically

    (vdu - udv)/v^2 yeah?

    When you work out udv, you need to differentiate (1+x)^1/2. When you do that, you get 0.5(1+x)^-1/2. In order to get rid of that negative power, they multiply the numerator and denominator by (1+x)^1/2, so the fraction stays the same, but it becomes more simplified.
 
 
 
  • 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
    What newspaper do you read/prefer?
    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.