• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

How do you derive this? (2 Viewers)

Joined
Sep 20, 2010
Messages
2,225
Gender
Undisclosed
HSC
2012
Do you mean differentiate? If so, did you mean:

?

You need to differentiate both sides with respect to x, remembering that y is a function of x - i.e. so then apply this knowledge when you do:

- remembering to use chain rule/product rule where needed
 

panda15

Alligator Navigator
Joined
Feb 22, 2012
Messages
675
Gender
Male
HSC
2013
Are you sure that it's Extension 1? It looks more like an Extension 2 question.
 

enigma_1

~~~~ Miss Cricket ~~~~
Joined
Feb 27, 2013
Messages
4,281
Location
Lords
Gender
Female
HSC
2014
Yep extension 1 and sorryr yeah I meant differentiate
 

enigma_1

~~~~ Miss Cricket ~~~~
Joined
Feb 27, 2013
Messages
4,281
Location
Lords
Gender
Female
HSC
2014
WHOa ok I've never done this before, can you please show me the method?sorry
 
Joined
Sep 20, 2010
Messages
2,225
Gender
Undisclosed
HSC
2012
If this is Ext1, you will need to rearrange the equation so that y is the subject.
 

panda15

Alligator Navigator
Joined
Feb 22, 2012
Messages
675
Gender
Male
HSC
2013
WHOa ok I've never done this before, can you please show me the method?sorry
Yeah the method that asianese showed is implicit differentiation, which only comes up in Extension 2.
 

enigma_1

~~~~ Miss Cricket ~~~~
Joined
Feb 27, 2013
Messages
4,281
Location
Lords
Gender
Female
HSC
2014
I can't make y the subject, it comes out as like x^2= (7-y^2)/(1-y)
 
Joined
Sep 20, 2010
Messages
2,225
Gender
Undisclosed
HSC
2012
Actually you can't make y the subject lol. Leave the question since it is for MX2 only. Yeah lol dw about it.
 
Joined
Sep 20, 2010
Messages
2,225
Gender
Undisclosed
HSC
2012
I'll show you the method anyway - the process is called 'implicit differentiation' because y is 'implicitly' - (NOT explicitly - y is not the subject here!) defined as a function of x. (We usually assume y is a function of x anyway)

So in the normal way, we can apply the 'differentiation operator', to both sides of the function. Our goal is to find .

Now we have

and differentiating term by term, we get that

. Now, we consider the first term in y^2 - use the chain rule. Remember if we had something like , to differentiate, bring down the power, then multiply by the derivative of the inside function. Do the same with y^2, except the 'inside function' is actually y itself, so its derivative is indeed . Repeat this process with the second term in x^2y, but remember this will have to ulilise the product rule - x^2 is multiplied by y, as well as a chain rule, since y is a function of x.

Once we have all our s appearing, collect all these on one side and rearrange to produce the result.

The final answer should be
 
Last edited:

enigma_1

~~~~ Miss Cricket ~~~~
Joined
Feb 27, 2013
Messages
4,281
Location
Lords
Gender
Female
HSC
2014
Wow thanks man!! So you're certaini that this wont be tested in MX1 exams yeah? It was in a MX1 textbook - odd..
 

QZP

Well-Known Member
Joined
Oct 7, 2013
Messages
839
Gender
Undisclosed
HSC
2014
What? I learnt implicit differentiation in year 11 MX1. It's not a MX2 technique :S
 

enigma_1

~~~~ Miss Cricket ~~~~
Joined
Feb 27, 2013
Messages
4,281
Location
Lords
Gender
Female
HSC
2014
Whaaaat now I'm really confused?! This thing was from Terry Lee.

There is actually a chapter on implicit diff in Cambridge but it says Extension or something.
@QZP, were you ever tested in an exam on it?
 

panda15

Alligator Navigator
Joined
Feb 22, 2012
Messages
675
Gender
Male
HSC
2013
What? I learnt implicit differentiation in year 11 MX1. It's not a MX2 technique :S
Your teacher probably taught it to you because it isn't a difficult concept, and it can help with some of the nastier 3U differentials, but it is a 4U technique. 3U will never get a question that involves implicit differentiation.
 
Joined
Sep 20, 2010
Messages
2,225
Gender
Undisclosed
HSC
2012
Whaaaat now I'm really confused?! This thing was from Terry Lee.

There is actually a chapter on implicit diff in Cambridge but it says Extension or something.
@QZP, were you ever tested in an exam on it?

Terry Lee has a habit of not reading syllabi when he writes textbooks. Ignore it.
 

QZP

Well-Known Member
Joined
Oct 7, 2013
Messages
839
Gender
Undisclosed
HSC
2014
Oh okay, I didn't know it was part of MX2. Thanks :)

@Hyper_bole
I wasn't tested on it, but honestly it's not all that hard to learn. And it makes the whole process of differentiation a lot cleaner (viewing d/dx as an operator vs. as a limiting process by first principles). I'd recommend knowing it whether or not it's in your test :p
 

Kiraken

RISK EVERYTHING
Joined
Jun 8, 2012
Messages
1,908
Gender
Undisclosed
HSC
N/A
What? I learnt implicit differentiation in year 11 MX1. It's not a MX2 technique :S
some people learn it in Extension 1 because as pointed out elsewhere in this thread it is in a chapter as an "extension" topic in the cambridge prelim 3u book. It is technically within mx2 but by that stage you should have the mathematical knowledge to understand it so you can learn it now but it is unlikely to be examinable in a 3u exam (although if your school makes it's own exams and you went through this topic at school it is best to confirm with the teacher if it will be examined)
 

anomalousdecay

Premium Member
Joined
Jan 26, 2013
Messages
5,766
Gender
Male
HSC
2013
Don't worry about it.

You will NEVER get this in a MX1 HSC paper.

It is too unfair.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top