Turn on thread page Beta
    • Thread Starter
    Offline

    11
    ReputationRep:
    Hi guys,
    I'm trying to find out dy/dx of the function on the attachment using implicit differentiation, apparently some of my answers to this have different sign that the text book answer i've done it over and over again but still can't find the stage that goes wrong in my working. would you please have a look at my working on the attachment and let me know if I'm wrong at a certain stage
    Attached Images
     
    Offline

    22
    ReputationRep:
    (Original post by Alen.m)
    Hi guys,
    I'm trying to find out dy/dx of the function on the attachment using implicit differentiation, apparently some of my answers to this have different sign that the text book answer i've done it over and over again but still can't find the stage that goes wrong in my working. would you please have a look at my working on the attachment and let me know if I'm wrong at a certain stage
    1. This would be easier done if you factorised out \cos y and used the product rule: \cos y(\cos x + \sin x) = \frac{1}{2}

    2. I'm not sure where your first term is coming from? What have you differentiated to get \cos x?
    Offline

    15
    ReputationRep:
    Hi, I cant open the image...
    what exam board are you doing?
    Offline

    22
    ReputationRep:
    (Original post by Alen.m)
    ...
    Given that I don't think what you're doing is currently right, I'll lead you down the right path:

    \cos y(\cos x+ \sin x) = \frac{1}{2}.

    Differentiate both sides w.r.t x:

    \displaystyle \frac{\mathrm{d}}{\mathrm{d}x}( \cos y (\cos x + \sin x)) = 0

    Product rule on the LHS:

    \displaystyle (\cos x + \sin x)\frac{\mathrm{d}}{\mathrm{d}x}  (\cos y) + \cos y\frac{\mathrm{d}}{\mathrm{d}x}(  \cos x + \sin x) = 0
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by Zacken)
    1. This would be easier done if you factorised out \cos y and used the product rule: \cos y(\cos x + \sin x) = \frac{1}{2}

    2. I'm not sure where your first term is coming from? What have you differentiated to get \cos x?
    i actually used product rule for the d/dx cosy cos x as i've shown in the attachment
    Attached Images
     
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by Student1408)
    Hi, I cant open the image...
    what exam board are you doing?
    AQA page 86 question 10 part b
    Offline

    22
    ReputationRep:
    (Original post by Alen.m)
    i actually used product rule for the d/dx cosy cos x as i've shown in the attachment
    \cos y - \sin y \frac{dy}{dx} does not mean cos y multiplied by sin y multiplied by dy/dx. You'll need to include brackets to make that a multiplication. As it stand, that's a subtraction.
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by Zacken)
    \cos y - \sin y \frac{dy}{dx} does not mean cos y multiplied by sin y multiplied by dy/dx. You'll need to include brackets to make that a multiplication. As it stand, that's a subtraction.
    can't find cos y- sin y which part of it you mean?
    Offline

    15
    ReputationRep:
    (Original post by Alen.m)
    Hi guys,
    I'm trying to find out dy/dx of the function on the attachment using implicit differentiation, apparently some of my answers to this have different sign that the text book answer i've done it over and over again but still can't find the stage that goes wrong in my working. would you please have a look at my working on the attachment and let me know if I'm wrong at a certain stage
    I said this sometime last week, you do not find dy/dx of a function.
    Here you could find dy/dx by differentiating with respect to x and then rearranging the find dy/dx. But you do not find dy/dx of a function - it doesn't make sense.
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by B_9710)
    I said this sometime last week, you do not find dy/dx of a function.
    Here you could find dy/dx by differentiating with respect to x and then rearranging the find dy/dx. But you do not find dy/dx of a function - it doesn't make sense.
    im not trying to find dy/dx here . I'm trying to differentiate cos y cos x+ cos y sin x =1/2 using product rule and then rearrange to find dy/dx
    Offline

    15
    ReputationRep:
    (Original post by Alen.m)
    im not trying to find dy/dx here . I'm trying to differentiate cos y cos x+ cos y sin x =1/2 using product rule and then rearrange to find dy/dx
    To find dy/dx, so you are trying to find dy/dx.

     \displaystyle \frac{d}{dx} (\cos y \cos x)=\sin y \sin x \frac{dy}{dx} .
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by B_9710)
    To find dy/dx, so you are trying to find dy/dx.

     \displaystyle \frac{d}{dx} (\cos y \cos x)=\sin y \sin x \frac{dy}{dx} .
    did you use product rule to find that?i use product rule but i keep getting wrong answer
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by B_9710)
    To find dy/dx, so you are trying to find dy/dx.

     \displaystyle \frac{d}{dx} (\cos y \cos x)=\sin y \sin x \frac{dy}{dx} .
    This is how i done it
    Attached Images
     
    Offline

    15
    ReputationRep:
    (Original post by Alen.m)
    did you use product rule to find that?i use product rule but i keep getting wrong answer
    My mistake

     \displaystyle \frac{d}{dx}(\cos y\cos x)=-(\cos x\sin y) \left (\frac{dy}{dx} \right ) -\sin x\cos y .
    I forgot to add the last bit.
    • Thread Starter
    Offline

    11
    ReputationRep:
    Name:  image.jpeg
Views: 73
Size:  380.0 KBName:  image.jpeg
Views: 73
Size:  380.0 KB
    (Original post by B_9710)
    My mistake

     \displaystyle \frac{d}{dx}(\cos y\cos x)=-(\cos x\sin y) \left (\frac{dy}{dx} \right ) -\sin x\cos y .
    I forgot to add the last bit.
    I've managed to diffrentatiate the whole equation but when it comes to rearenging to find dy/dx i just cant get the same answer as what the text book says . I've attached the text book answer and also the whole diffrentation that needs to be rearenged but i just cant get the write answer
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by Zacken)
    \cos y - \sin y \frac{dy}{dx} does not mean cos y multiplied by sin y multiplied by dy/dx. You'll need to include brackets to make that a multiplication. As it stand, that's a subtraction.
    I've managed to diffrentatiate the whole equation but when it comes to rearenging to find dy/dx i just cant get the same answer as what the text book says . I've attached the text book answer and also the whole diffrentation that needs to be rearenged but i just cant get the write answerAttachment 510085510087Name:  image.jpeg
Views: 76
Size:  218.9 KB
    Attached Images
     
    Offline

    18
    ReputationRep:
    (Original post by Alen.m)
    I've managed to diffrentatiate the whole equation but when it comes to rearenging to find dy/dx i just cant get the same answer as what the text book says . I've attached the text book answer and also the whole diffrentation that needs to be rearenged but i just cant get the write answerAttachment 510085510087Name:  image.jpeg
Views: 76
Size:  218.9 KB
    When you've completed the differentiation than you need to move everything that does not have DY/DX to the right hand side.

    Factor out DY/DX and then divide the right hand side by whats in the brackets.

    Posted from TSR Mobile
    Offline

    15
    ReputationRep:
    (Original post by Alen.m)
    I've managed to diffrentatiate the whole equation but when it comes to rearenging to find dy/dx i just cant get the same answer as what the text book says . I've attached the text book answer and also the whole diffrentation that needs to be rearenged but i just cant get the write answerAttachment 510085510087Name:  image.jpeg
Views: 76
Size:  218.9 KB
    Just take terms not containing dy/dx to the other side of the equation. Then take out a factor of dy/dx.
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by B_9710)
    Just take terms not containing dy/dx to the other side of the equation. Then take out a factor of dy/dx.
    well i know all of that but if you move (cos x- sin y dy/dx) to the other side, which sign will be changed like will it become -(cos x-sin y dy/dx)?
    Offline

    15
    ReputationRep:
    (Original post by Alen.m)
    well i know all of that but if you move (cos x- sin y dy/dx) to the other side, which sign will be changed like will it become -(cos x-sin y dy/dx)?
    Would it be beneficial for me to write it up on here?
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 7, 2016
The home of Results and Clearing

3,015

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
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.