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    Hi guys ,
    quick question what would your answer be to the picture i attached you here
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    What exactly are you trying to do?
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    Urhm, what exactly is the question?
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    That expression you wrote down would simplify to  \displaystyle 2xy \frac{dy}{dx} .
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     \displaystyle \frac{d}{dx} [f(y)] = \frac{dy}{dx} \frac{d}{dy} [f(y)] .
    It is almost as if the dy's cancel.
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    (Original post by B_9710)
    That expression you wrote down would simplify to  \displaystyle 2xy \frac{dy}{dx} .
    this is the part i don't understand . here's how i do it which gives me the wrong answer : d/dx of y^2 is equal to 2y and 2y. x would give you 2xy . i don't know where the dy/dx comes from
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    (Original post by Alen.m)
    this is the part i don't understand . here's how i do it which gives me the wrong answer : d/dx of y^2 is equal to 2y and 2y. x would give you 2xy . i don't know where the dy/dx comes from
    Actually  \displaystyle \frac{d}{dx} y^2 = 2y \frac{dy}{dx} .
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    Use the chain rule. y is defined in terms of x - it isn't just a constant. Thus you have to multiply through by dy/dx.

    (Original post by Alen.m)
    this is the part i don't understand . here's how i do it which gives me the wrong answer : d/dx of y^2 is equal to 2y and 2y. x would give you 2xy . i don't know where the dy/dx comes from
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    (Original post by Alen.m)
    this is the part i don't understand . here's how i do it which gives me the wrong answer : d/dx of y^2 is equal to 2y and 2y. x would give you 2xy . i don't know where the dy/dx comes from
    Have you covered the chain rule yet?
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    (Original post by B_9710)
    Actually  \displaystyle \frac{d}{dx} y^2 = 2y \frac{dy}{dx} .
    The original question was x*d/dx(y^2). Multiplying by the x gives 2xy*dy/dx, as required.
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    (Original post by Alen.m)
    this is the part i don't understand . here's how i do it which gives me the wrong answer : d/dx of y^2 is equal to 2y and 2y. x would give you 2xy . i don't know where the dy/dx comes from
    Because if you are being asked to differnetiate y^2 with respect to x (which is what d/dx is asking) you simply cannot do this as you cannot differentiate a function in respect to another function like that. So you have to implicity differnetiate where you use the forumla stated above.

    To give you y^2' *dy/dx *x=

    2xy dy/dx

    If it was asking to differentiate x^2 with respect to x you can do this and is a common result of 2x, hopefully that makes sense?
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    (Original post by constellarknight)
    The original question was x*d/dx(y^2). Multiplying by the x gives 2xy*dy/dx, as required.
    I know, I was replying to a question that the OP asked after the original post.
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    (Original post by Calzs34)
    Because if you are being asked to differnetiate y^2 with respect to x (which is what d/dx is asking) you simply cannot do this as you cannot differntiate a function in respect to another function like that. So you have to implicity differntiate where you use the forumla stated above.

    To give you y^2' *dy/dx *x

    If it was asking to differntiate x^2 with respect to x you can do this and is a common result of 2x, hopefully that makes sense?
    used the chain rule to solve it but still not getting it here's how i done it
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    (Original post by constellarknight)
    Use the chain rule. y is defined in terms of x - it isn't just a constant. Thus you have to multiply through by dy/dx.
    am i missing something here?
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    (Original post by lukejoshjames)
    Have you covered the chain rule yet?
    yes i did but i still get the answer as 2xy d/dx
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    (Original post by Alen.m)
    used the chain rule to solve it but still not getting it here's how i done it
    (Original post by Alen.m)
    am i missing something here?
    Try watching this video, you've made a mistake in the way you've differntiated, hopefully this clears it up:

    https://www.khanacademy.org/math/dif...ferentiation-1

    If you're still confused reply back, but it's better to work it out yourself imo
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    (Original post by Alen.m)
    yes i did but i still get the answer as 2xy d/dx
    You do not actually cancel the dy's. The  d/dy tells you to differentiate with respect to y and then you multiply through by dy/dx.
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    (Original post by B_9710)
    You do not actually cancel the dy's. The  d/dy tells you to differentiate with respect to y and then you multiply through by dy/dx.
    thanks man i got it now
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    (Original post by Calzs34)
    Try watching this video, you've made a mistake in the way you've differntiated, hopefully this clears it up:

    https://www.khanacademy.org/math/dif...ferentiation-1

    If you're still confused reply back, but it's better to work it out yourself imo
    the video was perfectly clear thanks
 
 
 
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