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    y = cosx/1-sinx , find dy/dx

    I know the answer is 1/1-sinx but I just can't get to it. Am I correct in thinking the quotient rule must be used?

    Thanks in advance.
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    (Original post by Outsanity)
    Am I correct in thinking the quotient rule must be used?
    no, you can write it as cosx(1-sinx)^{-1} and use the product and chain rules
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    (Original post by Pheylan)
    no, you can write it as cosx(1-sinx)^{-1} and use the product and chain rules
    Oh ok, that's fine, thanks

    Could you please tell me why you can't use the quotient rule? Is it the presence of the 1-sinx?
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    You can use it but what he suggested is much easier and it would be prudent to be able to spot this!
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    (Original post by Outsanity)
    Oh ok, that's fine, thanks

    Could you please tell me why you can't use the quotient rule? Is it the presence of the 1-sinx?
    i didn't say you can't use the quotient rule, either method will work
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    (Original post by Pheylan)
    i didn't say you can't use the quotient rule, either method will work
    Thanks for your help.
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    (Original post by boromir9111)
    You can use it but what he suggested is much easier and it would be prudent to be able to spot this!
    Wait, having used that i'm still having trouble.

    cosx(1-sinx)^-1 I differentiate to -cosx-cosx(1-sinx)^-2 -> cos^2x/1-sin^2x

    I think i'm going wrong somewhere but i'm not sure where?
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    (Original post by Outsanity)
    Wait, having used that i'm still having trouble.

    cosx(1-sinx)^-1 I differentiate to -cosx-cosx(1-sinx)^-2 -> cos^2x/1-sin^2x

    I think i'm going wrong somewhere but i'm not sure where?
    Use chain rule on (1-sin(x))^-1 and then product rule on cos(x)*(1-sin(x))^-1. Post further working if you get stuck
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    (Original post by boromir9111)
    Use chain rule on (1-sin(x))^-1 and then product rule on cos(x)*(1-sin(x))^-1. Post further working if you get stuck
    Argh i'm getting all mixed up in (1-sinx)^-3 which I know can't be right!

    So I have cosx(1-sinx)^-1

    Using the chain rules appears to give me -1 x cosx x (1-sinx)^-2 x -cosx which eventually cancels to 1, which obviously isn't right.

    Where am I going wrong here, have I missed a step?
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    (Original post by Outsanity)
    Argh i'm getting all mixed up in (1-sinx)^-3 which I know can't be right!

    So I have cosx(1-sinx)^-1

    Using the chain rules appears to give me -1 x cosx x (1-sinx)^-2 x -cosx which eventually cancels to 1, which obviously isn't right.

    Where am I going wrong here, have I missed a step?
    (1-sin(x))^-1 differentiated gives cos(x)*(1-sin(x))^-2

    this is your du/dx value.

    V = cos(x)

    v*du/dx = -sin(x)/1-sin(x)

    u*dv/dx = cos^(2)x/(1-sin^(2)x)
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    (Original post by boromir9111)
    (1-sin(x))^-1 differentiated gives cos(x)*(1-sin(x))^-2

    this is your du/dx value.

    V = cos(x)

    v*du/dx = -sin(x)/1-sin(x)

    u*dv/dx = cos^(2)x/(1-sin^(2)x)
    I was stating U = cos(x), don't know if that makes a difference?

    Anyway, so from that we deuce that Vdu/dx + Udv/dx =>

    -sinx/1-sinx + 1 but that doesn't come to 1/1-sinx I don't think?
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    (Original post by Outsanity)
    I was stating U = cos(x), don't know if that makes a difference?

    Anyway, so from that we deuce that Vdu/dx + Udv/dx =>

    -sinx/1-sinx + 1 but that doesn't come to 1/1-sinx I don't think?
    Shouldn't make a difference, answer comes out same, I was just doing it in my head quickly, so changed the symbols around!

    You are indeed correct, the negative will always be there. Either the answer in the book is wrong or my method is wrong somewhere!
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    (Original post by Outsanity)
    y = cosx/1-sinx , find dy/dx

    I know the answer is 1/1-sinx but I just can't get to it. Am I correct in thinking the quotient rule must be used?

    Thanks in advance.
    I prefer to use the quotient rule, but you don't have to. Use what you're comfortable with as the trickiest part is simplifying fully.

    The book is correct. Working is below:

    

y = \dfrac{\cos x}{1 - \sin x}





y' = \dfrac{(1 - \sin x)(-\sin x) - (\cos x)(-\cos x)}{(1 - \sin x)^2}



= \dfrac{(-\sin x + \sin^2 x) +  (\cos^2 x)}{(1 - \sin x)^2}



= \dfrac{(1 - \sin x)}{(1 - \sin x)(1 - \sin x)}



= \dfrac{1}{1 - \sin x}

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    Lol.....silly error made again!

    I canceled down early, if you multiply by 1-sin(x) to get the same fraction, then you get the same result from my method as well!
 
 
 
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