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Stationary points of...... Watch

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    cos(x+y)

    If I do df/dx and df/dy I get the same answer (-sin(x+y)) for both. Then if this has to equal zero then x+y has to equal zero which suggests that there are infinite values of x and y where there are stationary points. Surely this can't be right????


    Dan
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    Just thinking about it, have I messed up the differentiation of cos(x+y)??? I can't understand why -sin(x+y) would be wrong though?
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    Anyone??? :P
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    Use the compound angle formula and then differentiate
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    (Original post by danhirons)
    Just thinking about it, have I messed up the differentiation of cos(x+y)??? I can't understand why -sin(x+y) would be wrong though?
    Use the fact that cos(x+y) is the same as (cosx)(cosy) - (sinx)(siny); then you can differentiate with product rule
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    (Original post by chembob)
    Use the fact that cos(x+y) is the same as (cosx)(cosy) - (sinx)(siny); then you can differentiate with product rule

    As I'm doing partial differentiation though I end up with df/dx and df/dy as the same thing. Therefore I always end up with sin(x+y) = 0 leading to an infinite number of answers of x and y so how can I solve it when it's like this?
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    Yes. You end up with something like a ridge.
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    (Original post by SimonM)
    Yes. You end up with something like a ridge.
    So I can't actually calculate any stationary points for it? Do I just leave the question at this point then...
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    (Original post by danhirons)
    So I can't actually calculate any stationary points for it? Do I just leave the question at this point then...
    Of course you can:

    Any solutions to sin(x+y) = 0 is a stationary point. Therefore we have a series of lines of stationary points:

    x+y = 0
    x+y = pi
    ...
    x+y = n pi
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    (Original post by SimonM)
    Of course you can:

    Any solutions to sin(x+y) = 0 is a stationary point. Therefore we have a series of lines of stationary points:

    x+y = 0
    x+y = pi
    ...
    x+y = n pi

    Ahhhh thanks, I would rep you but for some reason it says I can't rep you again even though I'm pretty sure I haven't before :P !!!
 
 
 
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