Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    1
    ReputationRep:
    I am having a tough time with this question. i know i need to simplify it first, but how do i simplify ln[(x^3)-3x]?
    • Very Important Poster
    Offline

    21
    ReputationRep:
    Very Important Poster
    (Original post by yoshimatey)
    I am having a tough time with this question. i know i need to simplify it first, but how do i simplify ln[(x^3)-3x]?
    I don't think that you need to/can simplify it - either way I would just apply the chain rule straight away.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by SeanFM)
    I don't think that you need to/can simplify it - either way I would just apply the chain rule straight away.
    please could you walk me through that? thank you so much for your reply
    • Very Important Poster
    Offline

    21
    ReputationRep:
    Very Important Poster
    (Original post by yoshimatey)
    please could you walk me through that? thank you so much for your reply
    I could but that is against forum guidelines

    How can you use the chain rule on it?
    Offline

    3
    ReputationRep:
    (Original post by yoshimatey)
    I am having a tough time with this question. i know i need to simplify it first, but how do i simplify ln[(x^3)-3x]?
    You have to use the chain rule. Here is a video explaining it:
    http://www.examsolutions.net/maths-r...tutorial-1.php
    Basically what you would have to do it let y = ln(t)
    where t = x^3 - 3x (assuming that you have written it as I think you have i.e. everything is being naturally logged)
    Then you differentiate ln(t), which is 1 / t. You don't actually write t though you write what 't' is which is x^3 - 3x so you would get 1 / x^3 - 3x. Then (in the same line) you need to multiply that result by the differential of t, so you multiply the 1 / x^3 - 3x by the differential of x^3 - 3x. You should be able to get to the answer from there, if not then just watch some videos on exam solutions for it to make sense as it seems that you probably haven't learnt it so what I just said will probably make little sense
    Offline

    3
    ReputationRep:
    (Original post by yoshimatey)
    please could you walk me through that? thank you so much for your reply
    If you differentiate ln[x] you get 1/x

    If you differentiate ln[f(x)] you get f'(x) / f(x)

    In this case f(x) = (x^3)-3x

    f'(x) = ???

    So the derivative of ln[(x^3)-3x] = ??? / [(x^3)-3x], where you may be able to simplify afterwards (haven't checked)...
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by SeanFM)
    I could but that is against forum guidelines

    How can you use the chain rule on it?
    oh, i didn't realise. why is it against the rules?
    • Very Important Poster
    Offline

    21
    ReputationRep:
    Very Important Poster
    (Original post by yoshimatey)
    oh, i didn't realise. why is it against the rules?
    http://www.thestudentroom.co.uk/showthread.php?t=403989
    Offline

    3
    ReputationRep:
    (Original post by yoshimatey)
    x
    Just remember that the chain rule is for differentiating a function of a function.

    In general, you could have g[f(x)] where g and f could be any functions (such as polynomials, trigonometric, exponential, natural logs, etc.)

    To find the derivative of the expression, you would:
    1) find the derivative of g (the general thing) with respect to the thing inside (treating it like normal, ignoring the inside for now)
    2) find the derivative of f (the inside thing) with respect to x
    3) multiply these things together

    For example if you have sin[(x^3)-3x]:
    1) First find the derivative of the general thing. Derivative of sin is cos, so you would have cos[(x^3)-3x]
    2) But it's not this simple since you have a function inside the sin, so you have to deal with this as well. Derivative of the inside is (3x^2)-3
    3) Multiply them together to get [(3x^2)-3] * cos[(x^3)-3x]

    Another example... [sin(x)]^2
    1) The general thing is a quadratic if you ignore the thing inside. The derivative of a quadratic such as x^2 is 2x. So the first part of this would be 2*sin(x)
    2) Now differentiate the thing inside, to get cos(x)
    3) Answer is cos(x) * 2sin(x)

    In the case of the OP, the general thing was a natural log if you ignored the inside part. But then you had to account for the inside part afterwards.

    To understand/visualise why it works like this, I would recommend searching "chain rule khanacademy" on YouTube. I think he explains/derives these things in an intuitive way.

    Hope this helps
    Offline

    1
    ReputationRep:
    (3x^2-3)/(x^3-3x)

    Posted from TSR Mobile
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 26, 2015
Poll
Do I go to The Streets tomorrow night?
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.