Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    1
    ReputationRep:
    apparently, 1/2(e1+e2)^2 differentiates to e1+e2

    is this always the case?

    e.g would 0.3(e1+e2)^2 differentiate to 0.6e1 + 0.6e2

    or would 0.7(2e1+8e2)^2 differentiate to 1.4(2e1+8e2) = 2.8e1 +11.2e2?

    thanks
    Offline

    0
    ReputationRep:
    apparently, 1/2(e1+e2)^2 differentiates to e1+e2

    I don't know what you mean by e1 and e2, but this requires the chain rule.

    Let's say (e1+e2) = u, then 1/2(e1+e2)^2 = 1/2 x u^2

    Use dy/dx = dy/du x du/dx (y in terms of x)

    So dy/du = u and du/dx = d(e1+e2)/dx

    Say you have 1/2(x+1)^2 , this differentiates to x+1 as x+1 differentiates to 1.

    If you have 1/2(2x+1)^2, this differentiates to 2(2x+1)
    Offline

    2
    ReputationRep:
    or another way of doing it, as i don't really like the chain rule:

    times by the power
    take one from the power
    times by the differential of the things in brackets

    for example- 3(2x^2 +1)^3
    goes to 36x(2x^2 +1)^2 (i hope, i did have to go and choose a complicated example!)
    • Thread Starter
    Offline

    1
    ReputationRep:
    sorry for not making it clearer, think of e1 and e2 as "x" and "y"... is what i wrote correct or false?
    Offline

    2
    ReputationRep:
    (Original post by Aleeece123)
    or another way of doing it, as i don't really like the chain rule:

    times by the power
    take one from the power
    times by the differential of the things in brackets

    for example- 3(2x^2 +1)^3
    goes to 36x(2x^2 +1)^2 (i hope, i did have to go and choose a complicated example!)
    But that is the chain rule. :p:

    (Original post by redkopite)

    sorry for not making it clearer, think of e1 and e2 as "x" and "y"... is what i wrote correct or false?
    I don't know. Your notation is a little confusing.
    What's in the brackets? What are you finding?

    a(bx + c)^n

    Differentiates wrt x to:

    nab(bx + c)^n-1

    Because u = bx + c
    du/dx = b
    d/dx = du/dx * d/du

    But in your example, you say e1 and e2. If they are numbers, then they have no derivative. If they are terms of x, then yes, you are right. Except for your last one. It doesn't really make sense.

    Then you said, think of it as x and y. Well that would be:

    a(bx + cy)^n

    Differentiates wrt x to:

    na(b +c(dy/dx))(bx + cy)^n-1

    Through implicit differentiation.

    I think I've made this look very complicated to you, but you're not being very clear. I think my first example answers your question though.
    Offline

    2
    ReputationRep:
    (Original post by AnonyMatt)
    But that is the chain rule. :p:
    so it is... ahh that would explain a lot! Maybe i just don't like saying "chain rule" :p:
 
 
 
  • 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
    Has a teacher ever helped you cheat?
    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

    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.