username4992520
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I can't understand the proof for question C) in the 2013 MAT. The proof seems to ignore the chain rule for some reason. Could anybody clarify why isn't the chain rule used?

Link to the paper:https://www.maths.ox.ac.uk/system/fi...nts/test13.pdf
Link to the solutions: https://www.maths.ox.ac.uk/system/fi...utions13_0.pdf
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RDKGames
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(Original post by vikomen)
I can't understand the proof for question C) in the 2013 MAT. The proof seems to ignore the chain rule for some reason. Could anybody clarify why isn't the chain rule used?

Link to the paper:https://www.maths.ox.ac.uk/system/fi...nts/test13.pdf
Link to the solutions: https://www.maths.ox.ac.uk/system/fi...utions13_0.pdf
You dont need to ?? Just determine f''(x) which doesnt need chain rule since the arguments of f' and g' are linear with coefficient of x being 1. Then just make a substitution from x to 2x.

You can use the chain rule if you really want but it will just waste space on your paper.

Otherwise you can also begin with f'(2x) and THEN differentiate, which would indeed use the chain rule, but the extra factor of 2 you get would cancel out anyway.
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username4992520
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(Original post by RDKGames)
You dont need to ?? Just determine f''(x) which doesnt need chain rule since the arguments of f' and g' are linear with coefficient of x being 1. Then just make a substitution from x to 2x.

You can use the chain rule if you really want but it will just waste space on your paper.

Otherwise you can also begin with f'(2x) and THEN differentiate, which would indeed use the chain rule, but the extra factor of 2 you get would cancel out anyway.
But using the chain rule doesn't work. Since f'(2x)=2*g(2x+1) => f''(2x)=4h((2x+1)-1)=4h(2x)
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RDKGames
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(Original post by vikomen)
But using the chain rule doesn't work. Since f'(2x)=2*g(2x+1) => f''(2x)=4h((2x+1)-1)=4h(2x)
You didnt apply the chain rule on the LHS.

Differentiating f'(2x) leaves you with 2f''(2x)
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RDKGames
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(Original post by vikomen)
But using the chain rule doesn't work. Since f'(2x)=2*g(2x+1) => f''(2x)=4h((2x+1)-1)=4h(2x)
Also how is

f'(2x) = 2*g(2x-1) ??

Its just f'(2x) = g(2x+1) because the question gives f'(x) = g(x+1)
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username4992520
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(Original post by RDKGames)
You didnt apply the chain rule on the LHS.

Differentiating f'(2x) leaves you with 2f''(2x)
This makes more sense, thanks.
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