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C4 Indices Question

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 :smile:
Reply 1
Set u = x^2 and use the chain rule dydx=dudxdydu\frac{dy}{dx} = \frac{du}{dx}\frac{dy}{du} :smile:
Reply 2
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 :smile:


Is the working on the image by you? or is that the mark scheme? It looks right to me...
Reply 3
sonofdot
Set u = x^2 and use the chain rule dydx=dudxdydu\frac{dy}{dx} = \frac{du}{dx}\frac{dy}{du} :smile:


Ah, it's late at night, what would be dy/du? :s-smilie:

EDIT: Got it, thanks :p:

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.
Reply 4
ArchedEdge
Ah, it's late at night, what would be dy/du? :s-smilie:

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...
Reply 5
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...


:tongue: ha
Reply 6
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 :smile:


You've gotta use the chain rule mate.


So bold.

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