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secondary 4 | A Maths
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LockB
LockB

secondary 4 chevron_right A Maths chevron_right Singapore

need help with this qn, pls explain too

Date Posted: 3 years ago
Views: 164

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Mr Ang
Mr Ang's answer
347 answers (A Helpful Person)
1st
Hope this helps eh, thanks.
LockB
LockB
3 years ago
why x^2 and x no need to be replaced tho
Mr Ang
Mr Ang
3 years ago
Because we replace the more "complicated terms" like 1st and 2nd derivatives - in order to show that it will lead to "y".

Also, the right-hand side "y", does not contain 1st and 2nd derivatives, hence these 2 were to be replaced, so that you could simplify with terms in "x".

Hope this clarifies. Else I would get back to you tomorrow eh. Thanks.
LockB
LockB
3 years ago
ohh, so for these proving questions, we only replace those non-x ones, like derivatives and y such that everything is in x?
Eric Nicholas K
Eric Nicholas K
3 years ago
The full method of solving without "substitution" is only covered in JC, so at the secondary level we just verify the validity of the statement
J
J
3 years ago
The whole idea here is to get to the RHS, which is y.

The idea is to simplify the LHS such that you get y on the RHS.

So we only need to rewrite those dy/dx and d²y/dx² by substituting their equivalent terms in x such that we can simplify everything since it is now in terms of x.
J
J
3 years ago
At A levels, you will learn something called implicit differentiation (it is based on the chain rule), which avoids the need for these substitutions most of the time.
J
J
3 years ago
Alternatively, you could start from the RHS and work your way to LHS.

It will be the reverse of the steps that Mr Ang has shown you.

But this is much harder as you will need to know how to rewrite the terms as terms involving dy/dx and d²y/dx²
Mr Ang
Mr Ang
3 years ago
Well, pardon me replying so late, had a busy day today.

Anyway, I believe Mr Eric and Mr J had also explained the more advanced proof would be covered in the A Levels eh.

Have a good evening everyone. :)
LockB
LockB
3 years ago
ok thx for helping me :)