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    derp
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    \displaystyle \frac{dy}{dx}=\cos x \frac{dy}{dz} \implies \frac{d^2 y}{dx^2} =\cos x \frac{d}{dx}\left(\frac{dy}{dz}\  right) - \sin x \frac{dy}{dz}

    What's \frac{d}{dx}\left(\frac{dy}{dz}\  right)?
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    (Original post by 0-))
    \displaystyle \frac{dy}{dx}=\cos x \frac{dy}{dz} \implies \frac{d^2 y}{dx^2} =\cos x \frac{d}{dx}\left(\frac{dy}{dz}\  right) - \sin x \frac{dy}{dz}

    What's \frac{d}{dx}\left(\frac{dy}{dz}\  right)?
    i'm not sure
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    (Original post by Pheylan)
    i'm not sure
    Let u=\frac{dy}{dz}

    \displaystyle \frac{d}{dx}\left(\frac{dy}{dz}\  right) = \frac{du}{dx}=\frac{du}{dz}\frac  {dz}{dx}=...
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    (Original post by 0-))
    Let u=\frac{dy}{dz}

    \displaystyle \frac{d}{dx}\left(\frac{dy}{dz}\  right) = \frac{du}{dx}=\frac{du}{dz}\frac  {dz}{dx}=...
    \frac{du}{dx} ? :dontknow:
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    (Original post by Pheylan)
    \frac{du}{dx} ? :dontknow:
    Why do you think that? Did you think about my post?

    I showed you that \frac{d}{dx}\left(\frac{dy}{dz}\  right)=\frac{du}{dz}\times \frac{dz}{dx}

    where u=\frac{dy}{dz}.

    What's \frac{du}{dz}? It's the derivative of \frac{dy}{dz} with respect to z which is \frac{d^2 y}{dz^2}

    What's \frac{dz}{dx}? You've worked this out previously so I'm sure you can do it again.
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    (Original post by 0-))
    What's \frac{dz}{dx}?
    cosx
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    (Original post by Pheylan)
    cosx
    Yes. So using my last two posts, what's \frac{d}{dx}\left(\frac{dy}{dz}\  right)?

    The way you're answering my questions makes it seem like you don't know what's going on with my posts. Let me know if you are understanding everything that I'm doing.
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    (Original post by 0-))
    Yes. So using my last two posts, what's \frac{d}{dx}\left(\frac{dy}{dz}\  right)?

    The way you're answering my questions makes it seem like you don't know what's going on with my posts. Let me know if you are understanding everything that I'm doing.
    well i understood the u=\frac{dy}{dz} part but i really don't know how to find \frac{d}{dx}(\frac{dy}{dz})
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    (Original post by Pheylan)
    well i understood the u=\frac{dy}{dz} part but i really don't know how to find \frac{d}{dx}(\frac{dy}{dz})
    Can you point out the exact place where you got lost in my working?
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    (Original post by 0-))
    Can you point out the exact place where you got lost in my working?
    not really, no
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    (Original post by Pheylan)
    not really, no
    Well then I can't help you anymore.

    I told you that \frac{d}{dx}\left(\frac{dy}{dz}\  right) is the product of two terms, one of which I gave to you and the other you worked out. Why can't you finish it?
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    (Original post by Pheylan)
    well i understood the u=\frac{dy}{dz} part but i really don't know how to find \frac{d}{dx}(\frac{dy}{dz})

    \frac{d}{dx}(\frac{dy}{dz})=\fra  c{d}{dz}(\frac{dy}{dz})\times\fr  ac{dz}{dx}

    =\frac{d^2y}{dz^2}\times\frac{dz  }{dx}
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    crazy maths people and your black magik
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    (Original post by 0-))
    Well then I can't help you anymore.

    I told you that \frac{d}{dx}\left(\frac{dy}{dz}\  right) is the product of two terms, one of which I gave to you and the other you worked out. Why can't you finish it?
    oh i didn't realise that

    so is \frac{dy}{dz}=cosx\frac{d^2y}{dz  ^2} ?
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    (Original post by Pheylan)
    oh i didn't realise that

    so is \frac{dy}{dz}=cosx\frac{d^2y}{dz  ^2} ?
    Yes.
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    (Original post by 0-))
    Yes.
    but now i have to differentiate that with respect to x :confused:
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    (Original post by Pheylan)
    but now i have to differentiate that with respect to x :confused:
    No. Look back at my first post.

    I should have metioned what your initial problem was. You gave this line

    \frac{d^2y}{dx^2}=cosx\frac{d^2y  }{dz^2}-sinx\frac{dy}{dz}

    where you've assumed that the derivative of dy/dz with respect to x is d^2y/dz^2. You should now be able to see that this is wrong.
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    (Original post by 0-))
    No. Look back at my first post.

    I should have metioned what your initial problem was. You gave this line

    \frac{d^2y}{dx^2}=cosx\frac{d^2y  }{dz^2}-sinx\frac{dy}{dz}

    where you've assumed that the derivative of dy/dz with respect to x is d^2y/dz^2. You should now be able to see that this is wrong.
    yeah i see it's wrong but now i have

    (Original post by 0-))
    \frac{d^2 y}{dx^2} =\cos x \frac{d}{dx}\left(\frac{dy}{dz}\  right) - \sin x \frac{dy}{dz}
    which becomes

    \frac{d^2y}{dx^2}=cosx\frac{d}{d  x}(cosx\frac{d^2y}{dz^2})-sinx\frac{dy}{dz}

    which becomes

    \frac{d^2y}{dx^2}=cosx(cosx(\fra  c{d}{dx}(\frac{d^2y}{dz^2}))-sinx\frac{d^2y}{dz^2})-sinx\frac{dy}{dz}

    :confused:
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    Sorry, the post where I just said 'yes' was wrong.

    It should be:

    \displaystyle \frac{d}{dx}\left(\frac{dy}{dz}\  right) = \cos x \frac{d^2 y}{d z^2}

    This should have been implied from my earlier posts. So now you have

    \displaystyle \frac{d^2 y}{dx^2} = \cos^2 x \frac{dy}{dz} - \sin x \frac{dy}{dz}
 
 
 
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