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    This is my second post today on second differential equations (so I apologise) but need help on this.

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    The pic shows the working out of the first half of the question incase anyone wants to know the question initially but the black box is where I'm focusing at for this post.

    First question, I understand where \frac {d}{du} (underlined in red) came from but I don't know what happened to it to the second line where the product rule was used.

    Secondly, on the second line where the product rule was used, x has been substituted in for \frac {dx}{du} (underlined in blue) but wasn't this meant be differentiated or am I missing something?
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    (Original post by ManLike007)
    This is my second post today on second differential equations (so I apologise) but need help on this.

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    The pic shows the working out of the first half of the question incase anyone wants to know the question initially but the black box is where I'm focusing at for this post.

    First question, I understand where \frac {d}{du} (underlined in red) came from but I don't know what happened to it to the second line where the product rule was used.

    Secondly, on the second line where the product rule was used, x has been substituted in for \frac {dx}{du} (underlined in blue) but wasn't this meant be differentiated or am I missing something?
    For the second line:

    The first product is the derivative of the x with respect to u (i.e. dx/du), with the dy/dx unchanged.

    The second product is the x unchanged with the deritivative of the dy/dx with respect to u.

    I.e. \displaystyle x\times \frac{d}{du}\left(\frac{dy}{dx} \right)

    And via the chain rules this equals

    \displaystyle x\times \frac{d}{dx}\left(\frac{dy}{dx} \right)\times \frac{dx}{du}

    \displaystyle = x\times \frac{d^2y}{dx^2}\times \frac{dx}{du}

    I think that covers both your questions - if not, let me know.
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    (Original post by ghostwalker)
    For the second line:

    The first product is the derivative of the x with respect to u (i.e. dx/du), with the dy/dx unchanged.

    The second product is the x unchanged with the deritivative of the dy/dx with respect to u.

    I.e. \displaystyle x\times \frac{d}{du}\left(\frac{dy}{dx} \right)

    And via the chain rules this equals

    \displaystyle x\times \frac{d}{dx}\left(\frac{dy}{dx} \right)\times \frac{dx}{du}

    \displaystyle = x\times \frac{d^2y}{dx^2}\times \frac{dx}{du}

    I think that covers both your questions - if not, let me know.
    Oh ok that clears everything up, thanks for your time!
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    ghostwalker I'm sorry to bother you again but I need help on this because the text book doesn't have an example for using a substitution of y=\frac{z}{x}

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    I got as far as \frac{d^{2}z}{dx^{2}} = \frac{d}{dx} \times \frac{dz}{dx}

    Therefore, \frac{d^{2}z}{dx^{2}} = \frac{d}{dx} (x \frac{dy}{dx} + y) but I don't know how to apply the product rule from here to get the correct answer.

    (I didn't see the point of making a new thread in all honesty)
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    (Original post by ManLike007)
    ghostwalker I'm sorry to bother you again but I need help on this because the text book doesn't have an example for using a substitution of y=\frac{z}{x}

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    I got as far as \frac{d^{2}z}{dx^{2}} = \frac{d}{dx} \times \frac{dz}{dx}
    Just correct that bit. Taking the derivative, is applying a function, so that should be

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

    Therefore, \frac{d^{2}z}{dx^{2}} = \frac{d}{dx} (x \frac{dy}{dx} + y) but I don't know how to apply the product rule from here to get the correct answer.

    (I didn't see the point of making a new thread in all honesty)
    To start, the derivative of the sum of two functions is the sum of the two derivatives, so this becomes:

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

    I'd like to see you have a go from there. If you can do it - excellent. If not, try and be specific about the issue you're having. Can you identify the two parts in the product? Are you sufficiently familiar with the product rule itself to apply it, without simplifying. Does one of the two parts, rather than the other, cause the problem? The more precisely you can define the issue, the easier you'll find you can get past it. Apologies, if those were all obvious to you.
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    (Original post by ghostwalker)
    Just correct that bit. Taking the derivative, is applying a function, so that should be

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



    To start, the derivative of the sum of two functions is the sum of the two derivatives, so this becomes:

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

    I'd like to see you have a go from there. If you can do it - excellent. If not, try and be specific about the issue you're having. Can you identify the two parts in the product? Are you sufficiently familiar with the product rule itself to apply it, without simplifying. Does one of the two parts, rather than the other, cause the problem? The more precisely you can define the issue, the easier you'll find you can get past it. Apologies, if those were all obvious to you.
    Apologies for the late reply

    I'm still getting the wrong answer, I can see that I need to apply the product rule for \frac {d}{dx} (x \frac {dy}{dx}) where this is familiar from the first question of this thread above and also apply the product rule for \frac {d}{dx} (y) but for this, I don't know how to differentiate \frac {d}{dx} if I'm not mistaken

    The issue I'm having is that I don't know how to differentiate generally things like \frac {dy}{dx}
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    (Original post by ManLike007)
    Apologies for the late reply

    I'm still getting the wrong answer, I can see that I need to apply the product rule for \frac {d}{dx} (x \frac {dy}{dx}) where this is familiar from the first question of this thread above
    OK.

    and also apply the product rule for \frac {d}{dx} (y) but for this, I don't know how to differentiate \frac {d}{dx} if I'm not mistaken
    I suspect the notation mistake in your previous post is a symptom of your problem.

    \frac {d}{dx} (y) does not require the product rule. It's simply the derivative of y with respect to x, i.e. dy/dx.

    It does NOT mean d/dx times y, but rather the derivative of y with respect to x.
    The d/dx can be considered as a function acting on y.

    d (sin x)/dx for example is the function "take the derivative with respect to x" acting on sin x, which would be cos x, as I'm sure you're aware.


    The issue I'm having is that I don't know how to differentiate generally things like \frac {dy}{dx}
    In this case if you need to differenetiate by the same variable again, i.e. by x, then you'd go to the next level of derivative:

    \displaystyle \dfrac{d}{dx}\left(\frac{dy}{dx}  \right)=\frac{d^2y}{dx^2}

    But suppose you need to differentiate by u. Then you'd need the chain rule:

    \displaystyle \dfrac{d}{du}\left(\frac{dy}{dx}  \right)=\dfrac{d}{dx}\left(\frac  {dy}{dx}\right)\times\frac{dx}{d  u}

    which equals

    \displaystyle \frac{d^2y}{dx^2}\times \frac{dx}{du}
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    (Original post by ghostwalker)
    OK.



    I suspect the notation mistake in your previous post is a symptom of your problem.

    \frac {d}{dx} (y) does not require the product rule. It's simply the derivative of y with respect to x, i.e. dy/dx.

    It does NOT mean d/dx times y, but rather the derivative of y with respect to x.
    The d/dx can be considered as a function acting on y.

    d (sin x)/dx for example is the function "take the derivative with respect to x" acting on sin x, which would be cos x, as I'm sure you're aware.




    In this case if you need to differenetiate by the same variable again, i.e. by x, then you'd go to the next level of derivative:

    \displaystyle \dfrac{d}{dx}\left(\frac{dy}{dx}  \right)=\frac{d^2y}{dx^2}

    But suppose you need to differentiate by u. Then you'd need the chain rule:

    \displaystyle \dfrac{d}{du}\left(\frac{dy}{dx}  \right)=\dfrac{d}{dx}\left(\frac  {dy}{dx}\right)\times\frac{dx}{d  u}

    which equals

    \displaystyle \frac{d^2y}{dx^2}\times \frac{dx}{du}
    Oh my god I don't know why but I was overthinking what \frac {d}{dx} was because this is FP2, my bad.

    I went and attempted the last three questions of the exercise that involved substitution and I got them right, I can't thank you enough because I spent hours on substitution yesterday.
 
 
 
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