Solve using quotient rule y=5x^3/(4x^2-3)^3

I have manage to differentiate the question but I’ve got completely confused when trying to simplify the answer.

I have factored out the common factor but I’ve got a little confused on the way as to what to do next

Reply 1

mqb2766

Original post

by KingRich

I have manage to differentiate the question but I’ve got completely confused when trying to simplify the answer.
I have factored out the common factor but I’ve got a little confused on the way as to what to do next

Can you post a pic of what youve done? It shouldnt be too bad if you factor out the common parts correctly.

Reply 2

KingRich

Original post

by mqb2766

Can you post a pic of what youve done? It shouldnt be too bad if you factor out the common parts correctly.

Sorry, i had started the message but got distracted and forgot click update

Reply 3

mqb2766

Original post

by KingRich

Sorry, i had started the message but got distracted and forgot click update

You can cancel the (4x^2-3)^2 on the numerator with (part of) the denominator. But your last term (4x^2-3)-8x^2 on the penultimate line going to the last line isnt correct. Its subtraction, not multiplication.

Reply 4

KingRich

Original post

by mqb2766

You mean like this?

I did that previously but leaving the other 4x²-3 didn’t seem right to me.

Reply 5

mqb2766

Original post

by KingRich

You mean like this?

I did that previously but leaving the other 4x²-3 didn’t seem right to me.

Yes, now simplify the [...] on the numerator.

Reply 6

KingRich

Original post

by mqb2766

Yes, now simplify the [...] on the numerator.

Sorry, I took a break and started reading last night.

So, expand like so, and simply fly by extracting the common factor -8x² and multiplying by 15x²?

So, that would be -120x⁴(4x²-3)/(4x²-3)⁴ ?

IMG_2973.jpeg101.6KB

Reply 7

mqb2766

Original post

by KingRich

Sorry, I took a break and started reading last night.

So, expand like so, and simply fly by extracting the common factor -8x² and multiplying by 15x²?
So, that would be -120x⁴(4x²-3)/(4x²-3)⁴ ?

Youre subtracting 8x^2, not multiplying by -8x^2.

Reply 8

DFranklin

As a side note, obviously if the question says "use the quotient rule", you have no choice, but if you have something of the form $\dfrac{d}{dx}\dfrac{f(x)}{g(x)^\alpha}$ (with $\alpha \neq 1$ ), it's generally easier to rewrite this as $\dfrac{d}{dx}f(x)g(x)^{-\alpha}$ and use the product rule.

[they're not really *that* different, but you don't end up with as much to cancel at the simplification stage].

Reply 9

mqb2766

Original post

by DFranklin

As a side note, obviously if the question says "use the quotient rule", you have no choice, but if you have something of the form $\dfrac{d}{dx}\dfrac{f(x)}{g(x)^\alpha}$ (with $\alpha \neq 1$ ), it's generally easier to rewrite this as $\dfrac{d}{dx}f(x)g(x)^{-\alpha}$ and use the product rule.
[they're not really *that* different, but you don't end up with as much to cancel at the simplification stage].

I was going to post this at end as its a simple way of justifying why you must have a ^4 on the denominator (after cancelling) rather than a ^6. So even if you dont do it this way, it explains #4.

Reply 10

KingRich

Original post

by mqb2766

Youre subtracting 8x^2, not multiplying by -8x^2.

Oh, poop!! that must have been where I’ve been going wrong!!

Okay, so:

Late reply: partner wanted to go shopping for her birthday

Reply 11

KingRich

Original post

by DFranklin

As a side note, obviously if the question says "use the quotient rule", you have no choice, but if you have something of the form $\dfrac{d}{dx}\dfrac{f(x)}{g(x)^\alpha}$ (with $\alpha \neq 1$ ), it's generally easier to rewrite this as $\dfrac{d}{dx}f(x)g(x)^{-\alpha}$ and use the product rule.
[they're not really *that* different, but you don't end up with as much to cancel at the simplification stage].

Is this A-Level? I don’t believe I’ve yet covered this in my current course or from my A-Level.

Edit: I just realised what is happening here, you’re just multiplying the denominator with the numerator which explains the negative power and why chain rule would work here. Which I now see why the chain rule is the go to, over using quotient rule.

I’ve just covered differentiating exponentials and inverse trigs. That’s as much as I’ve done in this calculus unit from the course.

As you’re both here, could you help me understand what is the fundamental of calculus is? I’ve briefly learned that it’s what connects differentiation with integration but I don’t know the full concept. :smile:

Reply 12

mqb2766

Original post

by KingRich

Oh, poop!! that must have been where I’ve been going wrong!!
Okay, so:

Late reply: partner wanted to go shopping for her birthday

Id have taken just the - out and left it factorised so
-15x^2(...)/(...)
but otherwise fine (apart from the denom being 4x^2 not 4x^3 - typo)

Reply 13

KingRich

Original post

by mqb2766

Id have taken just the - out and left it factorised so
-15x^2(...)/(...)
but otherwise fine.

So, leave as it is but just take out the negative. Thank you. It’s annoying sometimes when I advance a lot and then recall so many formulas but forget how to do the simple maths that I learned way back.

Reply 14

mqb2766

Original post

by KingRich

note the edit about the denom typo.

Reply 15

KingRich

Original post

by mqb2766

note the edit about the denom typo.

Oh, yes. Indeed a typo.

Thank you for pointing that out.

Reply 16

DFranklin

Original post

by KingRich

Edit: I just realised what is happening here, you’re just multiplying the denominator with the numerator which explains the negative power and why chain rule would work here. Which I now see why the chain rule is the go to, over using quotient rule.

That's not how I'd describe it.
Firstly, I'm just rewriting the division as a multiplication by the reciprocal. Which is a purely algebraic operation (basicaly GCSE level manipulation).
Secondly, in terms of differentiation rules, I'm using the product rule instead of the quotient rule. But both methods use the chain rule: explicitly, I'm using it to find $\dfrac{d}{dx} f(x)^{-\alpha}$ and you are using it to find $\dfrac{d}{dx} f(x)^\alpha$ .

[incidentally, this is why i added the caveat ("with $\alpha \neq 1$ ). In that case, I'm using the chain rule to find $\dfrac{d}{dx} f(x)^{-1}$ , while with the quotient rule you just need to find $\dfrac{d}{dx} f(x)$ (and so you don't have an additional chain rule to apply)].

As you’re both here, could you help me understand what is the fundamental of calculus is? I’ve briefly learned that it’s what connects differentiation with integration but I don’t know the full concept. :smile:

The fundamental theorem of calculus is that $\displaystyle \dfrac{d}{dx} \int_a^x f(t) \,dt = f(x)$ (for f continuous, although you don't really need to worry about that).

Or to put it in words, "you can find the integral of f(x) by finding a function g(x) with $\dfrac{d}{dx} g(x) = f(x)$ ". (*)

If this seems "odd", or tautological, it's because the first way we teach people about integration usually starts from (*), but that's not what integration "really" is.

Fundamentally, integration is about finding the area under a curve. That's how it's defined at university, and there's some fairly complex machinery involved in defining how we do this rigourously. So the fundamental theorem of calculus is saying "if we define g(t) to be the area under y = f(x) between x = a and x = t, then $\dfrac{dg}{dt} = f(t)$ ".

And it's that result that lets us say "if want to find the area under y = f(x) between x = a and x = b, it's enough to find a function g(x) with dg/dx = f(x) and calculate g(b) - g(a)".

[to put it another way, it's the fundamental theorem of calculus that underpins pretty much everything you know about how to integrate functions.]

(edited 1 year ago)