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    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
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    (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?
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    Hi, I cant open the image...
    what exam board are you doing?
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    (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
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    (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
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    (Original post by Student1408)
    Hi, I cant open the image...
    what exam board are you doing?
    AQA page 86 question 10 part b
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    (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.
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    (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?
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    (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.
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    (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
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    (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} .
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    (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
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    (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
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    (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.
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    (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
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    (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: 65
Size:  218.9 KB
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    (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: 65
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.

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    (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: 65
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.
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    (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)?
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    (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?
 
 
 
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