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

C4 Parametric Differentiation Watch

Announcements
    • Thread Starter
    Offline

    16
    ReputationRep:
    I learned in C3 that one version of the chain rule is \frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}. There was no proof for it, which is annoying, so if anyone can prove it with A level knowledge that would be helpful.

    Now in C4, I've been told that I can use the chain rule, but rearranged into the form \frac{dy}{dx}=\frac{dy}{du}/ \frac{dx}{du}. Can somone tell me how they arrived at this result. I know dividing is the same as multiplying by the reciprocal, but surely it can't be the same for derivatives? And if it is, then why didn't they stick to the result taught in C3?
    Offline

    17
    ReputationRep:
    (Original post by ViralRiver)
    I learned in C3 that one version of the chain rule is \frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}. There was no proof for it, which is annoying, so if anyone can prove it with A level knowledge that would be helpful.

    Now in C4, I've been told that I can use the chain rule, but rearranged into the form \frac{dy}{dx}=\frac{dy}{du}/ \frac{dx}{du}. Can somone tell me how they arrived at this result. I know dividing is the same as multiplying by the reciprocal, but surely it can't be the same for derivatives? And if it is, then why didn't they stick to the result taught in C3?
    The proof of

    \frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}

    is no different to the proof of

    \frac{a}{b}=\frac{a}{c} \times \frac{c}{b}

    Which I suppose answers both of your questions, it's not special just because it's a derivative.
    • Thread Starter
    Offline

    16
    ReputationRep:
    Hmm okay, my teacher said that you could think of it as the 'du's cancelling out, but that's not a proper mathematical explanation, which is what led me to this topic,
    Offline

    17
    ReputationRep:
    (Original post by ViralRiver)
    Hmm okay, my teacher said that you could think of it as the 'du's cancelling out, but that's not a proper mathematical explanation, which is what led me to this topic,
    If you've got a parabola, obviously

     y = x^2

    Parametrically,

     x = t
     y = t^2

    Using the  y = x^2 , you can quite easily derive

     \frac{dy}{dx} = 2x

    Using the parametric equations you get

     \frac{dx}{dt} = 1

     \frac{dy}{dt} = 2t

    Which, using

    \frac{dy}{dx}=\frac{dy}{dt} \times \frac{dt}{dx}

    produces

    \frac{dy}{dx}= 2t \times \frac{1}{1} = 2t

    Then using the prior relationship between x and t, you get

     \frac{dy}{dx} = 2x

    Since  x = t


    So you are essentially cancelling out the variable t.
    Offline

    17
    ReputationRep:
    Actually scrap that above, I just worked out a much nicer 'proof'.

    y=f(x) \Rightarrow \frac{dy}{dx} = f'(x)


    Parametrically.
    x=g(t) \Rightarrow \frac{dx}{dt} = g'(t)
    y=h(t) \Rightarrow \frac{dy}{dt} = h'(t)



    \frac{dy}{dx} = \frac{dy}{dt} \times  \frac{dt}{dx}

    R.H.S = h'(t) \times \frac{1}{g'(t)} = \frac{h'(t)}{g'(t)}

    Going back to the original equations you can see that since

    y=f(x)

    and

    y=h(t)

    f(x) = h(t)

    and since

     x = g(t)

    h(t) = f[g(t)]

    h'(t) = f'[g(t)] \times g'(t) (chain rule)


    Plugging that all back into

    \frac{dy}{dx} = \frac{h'(t)}{g'(t)}

    gives

     \frac{dy}{dx} = \frac{ f'[g(t)] \times g'(t)}{g'(t)}

    Cancelling the g'(t) terms gives

     \frac{dy}{dx} = f'[g(t)] = f'(x) = \frac{dy}{dx} = L.H.S
 
 
 
  • 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
    Will you be richer or poorer than your parents?
    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.