Turn on thread page Beta
    • Thread Starter
    Offline

    1
    ReputationRep:
    y=ln 0.2x

    y=ln z and z=0.2x

    (dy/dx)=1/z

    (dz/dy)=0.2

    (dy/dx)=(1/(0.2x))*0.2

    =0.2/(0.2x)

    =1/x

    Cheers
    Offline

    22
    ReputationRep:
    (Original post by Brainfrozen)
    y=ln 0.2x

    y=ln z and z=0.2x

    (dy/dx)=1/z

    (dz/dy)=0.2

    (dy/dx)=(1/(0.2x))*0.2

    =0.2/(0.2x)

    =1/x

    Cheers
    Yeah, that's correct. You can see it is because ln 0.2x = ln 0.2 + ln x = constant + ln x

    So the derivative of that is the same as the derivative of ln x.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Zacken)
    Yeah, that's correct. You can see it is because ln 0.2x = ln 0.2 + ln x = constant + ln x

    So the derivative of that is the same as the derivative of ln x.
    Cheers!
    Offline

    22
    ReputationRep:
    (Original post by Brainfrozen)
    Cheers!
    No problem.
    • Thread Starter
    Offline

    1
    ReputationRep:
    Now I have another problem and my assigned lecturer basically says he's too busy to help me, even although this module of the course is costing 500 pounds. Hardly had any contact with him and he says he's too busy marking assessments. Unreal.

    I have gotten as far as this;

    (2x + 1)^3 x 6x + (3x^2 - 1) x 6(2x + 1)^2

    It then says extract common factors, which I thought I knew what that meant but I'm confused by their result;

    6(2x + 1)^2 ((2x + 1)x + (3x^2 - 1))

    I'm not a stupid person, but with no support from my lecturer I feel stupid and depressed and with a huge possibility now of not getting pass the first module of this HNC - and I already have an HNC in the bag.
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by Brainfrozen)
    Now I have another problem and my assigned lecturer basically says he's too busy to help me, even although this module of the course is costing 500 pounds. Hardly had any contact with him and he says he's too busy marking assessments. Unreal.

    I have gotten as far as this;

    (2x + 1)^3 x 6x + (3x^2 - 1) x 6(2x + 1)^2

    It then says extract common factors, which I thought I knew what that meant but I'm confused by their result;

    6(2x + 1)^2 ((2x + 1)x + (3x^2 - 1))

    I'm not a stupid person, but with no support from my lecturer I feel stupid and depressed and with a huge possibility now of not getting pass the first module of this HNC - and I already have an HNC in the bag.
    6x(2x+1)^3+6(2x+1)^2(3x^2-1)

    Let a=2x+1 \Rightarrow 6xa^3+6a^2(3x^2-1)

    Factor the 6a^2

    6a^2(ax+(3x^2-1))

    Back substitute
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by RDKGames)
    6x(2x+1)^3+6(2x+1)^2(3x^2-1)

    Let a=2x+1 \Rightarrow 6xa^3+6a^2(3x^2-1)

    Factor the 6a^2

    6a^2(ax+(3x^2-1))

    Back substitute
    Thanks again mate, easy to understand when someone explains it properly. Why my lecturer couldn't spend two minutes doing that - I have no idea.
    • Thread Starter
    Offline

    1
    ReputationRep:
    I'm fine with converting between very simple primitives and derivatives, but can someone help me with this?

    change this primitive to function: -1/2x^2
    change this function to primitive: 1/x^2

    baby steps will be much appreciated.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: December 10, 2016
The home of Results and Clearing

2,263

people online now

1,567,000

students helped last year

University open days

  1. Sheffield Hallam University
    City Campus Undergraduate
    Tue, 21 Aug '18
  2. Bournemouth University
    Clearing Open Day Undergraduate
    Wed, 22 Aug '18
  3. University of Buckingham
    Postgraduate Open Evening Postgraduate
    Thu, 23 Aug '18
Poll
A-level students - how do you feel about your results?
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.