Differentiation AS maths

Announcements Posted on
How helpful is our apprenticeship zone? Have your say with our short survey 02-12-2016
    Offline

    2
    ReputationRep:
    (Original post by aliiceconnor)
    It's C1
    LOL and I got an A at C1. Keep up the good work. I would try be of more use but I see Sean is doing a great job as always and beat me to it.
    Online

    3
    ReputationRep:
    (Original post by aliiceconnor)
    So f'(-2)=-4

    -4=-4a+b-2a
    -8=-8a+2b-4a

    4a-2b+13=0
    -12a+2b+8=0

    4a+13-12a+8=0
    -8a+21=0
    f'(x) = 2ax + b, not 2ax + b - 2a. can you see why?
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by SeanFM)
    f'(x) = 2ax + b, not 2ax + b - 2a. can you see why?
    Not really,
    Would it be
    4a+13-8a+8=0
    So -4a+21=0?
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Epistemolog y)
    LOL and I got an A at C1. Keep up the good work. I would try be of more use but I see Sean is doing a great job as always and beat me to it.
    Thanks haha
    Online

    3
    ReputationRep:
    (Original post by aliiceconnor)
    Not really,
    Would it be
    4a+13-8a+8=0
    So -4a+21=0?
    I see a 13 in there so I see you're referring to the first equation you have set up which is correct.. but you are not taking in what I am saying about the a^2 that you put in there and how we are yet to set up the second equation before we can solve them simultaneously.

    We have that y = ax^2 + bx + a^2. When we differentiate this to find f'(x) you can differentiate individually ax^2, bx and a^2 * x^0 to give 2ax, b and 0, so f'(x) = 2ax + b + 0. Hence f'(-2) = 2 * a * -2 + b, and so f'(-2) = b - 4a and you know what f'(-2) is. Let me know if any of that is unclear
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by SeanFM)
    I see a 13 in there so I see you're referring to the first equation you have set up which is correct.. but you are not taking in what I am saying about the a^2 that you put in there and how we are yet to set up the second equation before we can solve them simultaneously.

    We have that y = ax^2 + bx + a^2. When we differentiate this to find f'(x) you can differentiate individually ax^2, bx and a^2 * x^0 to give 2ax, b and 0, so f'(x) = 2ax + b + 0. Hence f'(-2) = 2 * a * -2 + b, and so f'(-2) = b - 4a and you know what f'(-2) is. Let me know if any of that is unclear
    That's what I did, I sent the next step when I tried to solve it
    Online

    3
    ReputationRep:
    (Original post by aliiceconnor)
    That's what I did, I sent the next step when I tried to solve it
    If ever you set up two simultaneous equations and then solve then and then the solutions satisfy both equations (i.e put them back in to both equation 1 and 2), then assuming your equations are correct your answer is correct.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by SeanFM)
    If ever you set up two simultaneous equations and then solve then and then the solutions satisfy both equations (i.e put them back in to both equation 1 and 2), then assuming your equations are correct your answer is correct.
    Okay, thank you so much for all your help!
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: October 18, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Today on TSR
Poll
Would you rather have...?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

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
Study resources

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.