Turn on thread page Beta

AS-Level Maths - Differentiation help please! watch

    • Thread Starter
    Offline

    0
    ReputationRep:
    Hiya,

    I've just got down to starting some homework from maths and don't really understand how to do some of the questions for it. I've already got the answers in the back of the book, but I don't really know how they got to it and wanted to know if anyone could explain.

    1) Rewrite these equations using index notation, and then differentiate Y with respect to X.

    g) y= -(1/x^2)

    The answer in the book is 2/x^3 but i don't really know how they got to it. If anyone could help, thanks!

    The other question is this one:

    5) Given y=f(x) find f'(x) in each case:
    g) f(x) = ((x+3)^2) / x

    The answer is apparently 1 - (9/x^2) but again I don't really know how they got to it. If you could explain the step-by-step process of it at all, I would really appreciate it.

    Thanks in advance
    Offline

    2
    ReputationRep:
    1.
    g) rewritten in index notation is -1x^-2.
    This is then differentiated to 2x^-3 which is equivalent to 2/x^3

    5.
    g) expand brackets to get (x^2+6x+9)/x
    Which then becomes x+6+9x^-1 due to rules of indicies.
    Which is then differentiated to 1-9x^-2 or 1-(9/x^2) as your book says.

    Hope this helps.

    If you need more explanation I can help just didn't want to go into too much detail incase you understand it all already and just needed a pointer.
    Source: did AS last year and got an A
    Offline

    3
    ReputationRep:
    (Original post by Dubois98)
    Hiya,

    I've just got down to starting some homework from maths and don't really understand how to do some of the questions for it. I've already got the answers in the back of the book, but I don't really know how they got to it and wanted to know if anyone could explain.

    1) Rewrite these equations using index notation, and then differentiate Y with respect to X.

    g) y= -(1/x^2)

    The answer in the book is 2/x^3 but i don't really know how they got to it. If anyone could help, thanks!

    The other question is this one:

    5) Given y=f(x) find f'(x) in each case:
    g) f(x) = ((x+3)^2) / x

    The answer is apparently 1 - (9/x^2) but again I don't really know how they got to it. If you could explain the step-by-step process of it at all, I would really appreciate it.

    Thanks in advance
    Has your teacher thoroughly explained the concept of dy/dx (or "f'(x)")?

    If you can write y as a series of powers of x - for example,

    y = (ax^4) + (bx^3) + (cx^2) +(dx) + (e) + (fx^-1) + (gx^-2) + (hx^-3),

    then (noting that e = ex^0), the differential will be

    (4ax^3) + (3bx^2) + (2cx) + (1d) + (0) + (-1fx^-2) + (-2gx^-3) + (-3hx^-4).

    The manipulation is therefore straightforward provided you can write the equation for y in the appropriate format.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Bjc00)
    1.
    g) rewritten in index notation is -1x^-2.
    This is then differentiated to 2x^-3 which is equivalent to 2/x^3

    5.
    g) expand brackets to get (x^2+6x+9)/x
    Which then becomes x+6+9x^-1 due to rules of indicies.
    Which is then differentiated to 1-9x^-2 or 1-(9/x^2) as your book says.

    Hope this helps.

    If you need more explanation I can help just didn't want to go into too much detail incase you understand it all already and just needed a pointer.
    Source: did AS last year and got an A

    Thanks for the response - I understand the first question, but am still a bit stumped on the 2nd. Why does it become x+6+9x^-1? Should it not be something more like 2x+6+0? or am I completely at the wrong end of the stick?
    Offline

    2
    ReputationRep:
    (Original post by Dubois98)
    Thanks for the response - I understand the first question, but am still a bit stumped on the 2nd. Why does it become x+6+9x^-1? Should it not be something more like 2x+6+0? or am I completely at the wrong end of the stick?
    To make it into an easier form to deal with get it all on 1 line. When x^2+6x+9 is all divided by x the powers are taken away from each other. x is x^1 and so x^2 becomes x^1 (or just x), 6x becomes 6x^0 (or just 6) and the 9 becomes 9x^-1. then differentiate x+6+9x-1.
    Hope that's easier to see now.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: September 23, 2014
The home of Results and Clearing

2,761

people online now

1,567,000

students helped last year

University open days

  1. Bournemouth University
    Clearing Open Day Undergraduate
    Wed, 22 Aug '18
  2. University of Buckingham
    Postgraduate Open Evening Postgraduate
    Thu, 23 Aug '18
  3. University of Glasgow
    All Subjects Undergraduate
    Tue, 28 Aug '18
Poll
How are you feeling about GCSE results day?
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

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.