Can’t find the derivative of this

[Jesse_551]

What is the Derivative of this function?

y=ln(4-2x^3)

[Matthew_Whelan]

y=ln⁡〖(4-2x^3 〗)
let u=4-2x^3
y=ln⁡u
□(24&dy)/□(24&dx)=dy/du×du/dx
□(24&dy)/□(24&dx)=1/u×-6x^2
□(24&dy)/□(24&dx)=(-6x^2)/(4-2x^3 )

(Edit: I don't know how to make it look nice on AN, but you use the chain rule to attain the answer).

[Nature]

y=ln⁡〖(4-2x^3 〗)
let u=4-2x^3
y=ln⁡u
□(24&dy)/□(24&dx)=dy/du×du/dx
□(24&dy)/□(24&dx)=1/u×-6x^2
□(24&dy)/□(24&dx)=(-6x^2)/(4-2x^3 )

(Edit: I don't know how to make it look nice on AN, but you use the chain rule to attain the answer).

https://atarnotes.com/forum/index.php?topic=165627.0 - LaTeX Guide

[DrDusk]

y=ln⁡〖(4-2x^3 〗)
let u=4-2x^3
y=ln⁡u
□(24&dy)/□(24&dx)=dy/du×du/dx
□(24&dy)/□(24&dx)=1/u×-6x^2
□(24&dy)/□(24&dx)=(-6x^2)/(4-2x^3 )

(Edit: I don't know how to make it look nice on AN, but you use the chain rule to attain the answer).

Just curious as to why your doing it this way?

[Bri MT]

Just curious as to why your doing it this way?

In my experience, most vic students aren't taught this and don't learn it.

If people wish to use it for vce methods it can be useful but make sure you don't lose working marks from it.

[colline]

Matthew_whelan’s method is the standard one taught in VIC but you will not lose working out marks for using DrDusk’s method on the exam.

It’s a similar case with the chain rule as well where textbooks make you write ‘let u = ___’ etc and while it’s useful to understand the logic behind it, it’s incredibly time consuming for the exam, not to mention unnecessary.

[Bri MT]

Matthew_whelan’s method is the standard one taught in VIC but you will not lose working out marks for using DrDusk’s method on the exam.

It’s a similar case with the chain rule as well where textbooks make you write ‘let u = ___’ etc and while it’s useful to understand the logic behind it, it’s incredibly time consuming for the exam, not to mention unnecessary.

To clarify, my comment wasn't at all "don't use the fast method" it was meant as "if you're doing the fast method still show working rather than just doing it in your head" (something I was guilty of too often)

[DrDusk]

In my experience, most vic students aren't taught this and don't learn it.

That's so bizarre. I would assume if students are taught that d/dx(lnx) = 1/x then they would also be taught it's standard form as both are the exact same thing.

[Sine]

The method below was the way I always did log derivatives in VIC probably just depends on the teacher. I'm 100% certain VCE textbooks also show this way somewhere in them.



Sure, you would teach the chain rule initially but once students have the hang of that you would give them the shortcut or they will work out the shortcut for themselves.

I think if a question is 1-2 marks you should be fine going with the shortcut but if it's 3 they are probably implying to show the chain rule.

[SmellsLikeTeenSpirit]

Personally, I prefer Matthew's approach. It is so much more applicable at the day of VCE exam:
1) The basic formula for the chain rule and the derivative of ln(x) are readily accessible at the exam, being given at the end of exam papers.
2) I could not find in Mathsquest 12 that formula proposed by Dr Dusk. Its differential version is derived in Section 9.2, but it is a bit too optimistic to expect that a student can recollect (or derive) it under the exam stress and duress.

[DrDusk]

Personally, I prefer Matthew's approach. It is so much more applicable at the day of VCE exam:
1) The basic formula for the chain rule and the derivative of ln(x) are readily accessible at the exam, being given at the end of exam papers.
2) I could not find in Mathsquest 12 that formula proposed by Dr Dusk. Its differential version is derived in Section 9.2, but it is a bit too optimistic to expect that a student can recollect (or derive) it under the exam stress and duress.

I don't get how it is optimistic. For NSW students it's common knowledge and it's very easy to remember. As for the derivation, it's literally just this



This literally boils down to what Mathew said in that d/du(lnu) = 1/u, which the VCE students are already being taught.

There's nothing about this that's being too optimistic. If everyone in NSW can easily remember it I don't see any disadvantage of it being used in VCE. They can easily remember it too. It's saving soo much unnecessary working out and time as well =)

[colline]

To clarify, my comment wasn't at all "don't use the fast method" it was meant as "if you're doing the fast method still show working rather than just doing it in your head" (something I was guilty of too often)

Oh yes sorry! I didn’t mean to make my reply come off like that. Of course, working out is the most important thing on the exam and that’s where most of the marks are.

Personally, I prefer Matthew's approach. It is so much more applicable at the day of VCE exam:
1) The basic formula for the chain rule and the derivative of ln(x) are readily accessible at the exam, being given at the end of exam papers.
2) I could not find in Mathsquest 12 that formula proposed by Dr Dusk. Its differential version is derived in Section 9.2, but it is a bit too optimistic to expect that a student can recollect (or derive) it under the exam stress and duress.

Is it just me who thinks the non-chain rule method is SO much easier? 
Also why would it be more applicable for the exam? From reading examiners reports, I can’t find anywhere where you’d actually be given marks for using the chain rule as opposed to the “shortcut”

[milanander]

The method below was the way I always did log derivatives in VIC probably just depends on the teacher. I'm 100% certain VCE textbooks also show this way somewhere in them.

Maybe the study design has changed since you did methods? That formula is not in the mathsquest textbook and I don't think it's in cambridge either.

Somewhat related question: for product rule / quotient rule, in the textbook and pretty much everything I find from VCAA the working out says you have to "define u and v as functions of x" so for example


so from the discussion in this thread am I correct to assume that I don't actually need to write all that? can anyone who has done methods pls confirm? thanks.

[Sine]

Maybe the study design has changed since you did methods? That formula is not in the mathsquest textbook and I don't think it's in cambridge either.

Somewhat related question: for product rule / quotient rule, in the textbook and pretty much everything I find from VCAA the working out says you have to "define u and v as functions of x" so for example


so from the discussion in this thread am I correct to assume that I don't actually need to write all that? can anyone who has done methods pls confirm? thanks.

I completed methods with the current study design.

The formula I wrote isn't the EXACT formula within the textbook but basically the same idea is there in the Cambridge book (bottom of page 387). I don't know about maths quest tho.

[S_R_K]

Given y = ln(4 – 2x^3), and needing to find the derivative, it's just a pointless ritualistic exercise in bookkeeping to expect students to write down: "Let u = ..., then dy/dx = dy/du, blah blah blah". It's obvious that the derivative is –6x^2 / (4 – 2x^3).

Similarly for product and quotient rule. Given y = e^x * (x+1)^2, and needing to find the derivative, there's no need to write u = ... and v = ... The first line can be e^x * 2(x + 1) + e^x * (x + 1)^2.

(Notice that the previous example also highlights the silliness of requiring students to write down all the substitutions, because arguably students should write down a third substitution w = ... when chain-ruling (x+1)^2).