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    Just need someone to explain this question from the 2007 Jan paper.

    I don't get part b) of the question, and how you can use part a) to help you figure part b out.

    Does it simply become e^((x^2)ln2)

    Could someone please explain?

    Thanks in advance
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    Set u = x^2 and use the chain rule \frac{dy}{dx} = \frac{du}{dx}\frac{dy}{du}
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    (Original post by ArchedEdge)
    Just need someone to explain this question from the 2007 Jan paper.

    I don't get part b) of the question, and how you can use part a) to help you figure part b out.

    Does it simply become e^((x^2)ln2)

    Could someone please explain?

    Thanks in advance
    Is the working on the image by you? or is that the mark scheme? It looks right to me...
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    (Original post by sonofdot)
    Set u = x^2 and use the chain rule \frac{dy}{dx} = \frac{du}{dx}\frac{dy}{du}
    Ah, it's late at night, what would be dy/du?

    EDIT: Got it, thanks :p:

    (Original post by josh_a_y)
    Is the working on the image by you? or is that the mark scheme? It looks right to me...
    It's the markscheme.
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    (Original post by ArchedEdge)
    Ah, it's late at night, what would be dy/du?
    Late at night is no excuse for a potential Caian! :whip: If you set u=x^2, y=2^u, so dy/du is just what you found in part a
    edit: too slow...
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    (Original post by sonofdot)
    Late at night is no excuse for a potential Caian! :whip: If you set u=x^2, y=2^u, so dy/du is just what you found in part a
    edit: too slow...
    ha
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    (Original post by ArchedEdge)
    Just need someone to explain this question from the 2007 Jan paper.

    I don't get part b) of the question, and how you can use part a) to help you figure part b out.

    Does it simply become e^((x^2)ln2)

    Could someone please explain?

    Thanks in advance
    You've gotta use the chain rule mate.


    So bold.
 
 
 
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