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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Damo17 on December 11, 2008, 03:49:13 pm

Title: Calculus question....
Post by: Damo17 on December 11, 2008, 03:49:13 pm: Differentiate with respect to x:
$y= 4 \sqrt{cos(x+1)}$

Tried it but unsure of what to let u equal.
Title: Re: Calculus question....
Post by: ell on December 11, 2008, 03:56:02 pm: Let $u=\cos(x+1)$ , so then $y=4u^{\frac{1}{2}}$
Title: Re: Calculus question....
Post by: shinny on December 11, 2008, 03:56:41 pm: What do you mean what to let 'u' equal? Hopefully you're not confusing diff with anti-diff here...normally I'd do this without having to define anything and just go straight for the chain rule in 1 line, but if you must, let cos(x+1) be u then and then chain rule it.
Title: Re: Calculus question....
Post by: Collin Li on December 11, 2008, 03:57:06 pm: Let $u=\cos{(x+1)}$

$\implies y = 4\sqrt{u}$

$\frac{du}{dx} = -\sin{(x+1)}$ (You can use chain rule to find this, if you can't see it)

$\frac{dy}{du} = \frac{2}{\sqrt{u}}$

$\implies \frac{dy}{dx} = \frac{-2\sin{(x+1)}}{\sqrt{\cos{(x+1)}}}$
Title: Re: Calculus question....
Post by: Damo17 on December 11, 2008, 03:58:37 pm: thanks all, i got it now.
Title: Re: Calculus question....
Post by: Damo17 on December 11, 2008, 04:00:24 pm: Quote from: shinjitsuzx on December 11, 2008, 03:56:41 pm
What do you mean what to let 'u' equal? Hopefully you're not confusing diff with anti-diff here...normally I'd do this without having to define anything and just go straight for the chain rule in 1 line, but if you must, let cos(x+1) be u then and then chain rule it.

I would normally do that but my teacher has given us questions that we have to use the long way and questions to do it the quick way.
Title: Re: Calculus question....
Post by: Damo17 on December 11, 2008, 09:26:31 pm: Another question using the product rule.

Differentiate:
$y= ({x^2}-1)^2(x+5)^2$

I can do most of it but hit a road block. :(
Title: Re: Calculus question....
Post by: Collin Li on December 11, 2008, 09:34:12 pm: Let $u = (x^2-1)^2$ and $v = (x+5)^2$

$\implies u' = 4x(x^2-1)$ and $v' = 2(x+5)$

$\implies \frac{dy}{dx} = 4x(x^2-1)(x+5)^2 + 2(x+5)(x^2-1)^2$

$= 2(x^2-1)(x+5)[2x(x+5) + (x^2-1)]$

$= 2(x^2-1)(x+5)(3x^2 + 10x - 1)$
Title: Re: Calculus question....
Post by: Damo17 on December 11, 2008, 09:37:21 pm: ^
Thanks coblin,
I forgot to change the 4x to 2x!!
Title: Re: Calculus question....
Post by: Synesthetic on December 11, 2008, 11:38:19 pm: Quote from: shinjitsuzx on December 11, 2008, 03:56:41 pm
What do you mean what to let 'u' equal? Hopefully you're not confusing diff with anti-diff here...normally I'd do this without having to define anything and just go straight for the chain rule in 1 line, but if you must, let cos(x+1) be u then and then chain rule it.

Integration by u-substitution isn't in Methods anyway [as in the linear substitution form, which is in Specialist], whereas chain rule by this technique is in almost all the textbooks. (And I agree, I don't think the latter helps much!)
Title: Re: Calculus question....
Post by: shinny on December 11, 2008, 11:44:12 pm: Ahah yeh didn't realise it was in the methods board; woops.
Title: Re: Calculus question....
Post by: Mao on December 12, 2008, 12:11:41 am: Damo still need to be careful about it though, assuming if he is doing Spesh and MUEP Math next year =)

you should learn to do the Chain Rule without using a u-substitution. Whilst it doesn't change the answer, it can over-complicate things and make your workings unnecessarily big, it is almost necessary when you are in spesh and uni-math unless you can fit a lot of workings in a small space.
Title: Re: Calculus question....
Post by: Damo17 on December 12, 2008, 01:02:38 pm: Quote from: Mao on December 12, 2008, 12:11:41 am
Damo still need to be careful about it though, assuming if he is doing Spesh and MUEP Math next year =)

you should learn to do the Chain Rule without using a u-substitution. Whilst it doesn't change the answer, it can over-complicate things and make your workings unnecessarily big, it is almost necessary when you are in spesh and uni-math unless you can fit a lot of workings in a small space.
Thanks for the advice Mao.

I can do most question without u-substitution but there are a few that I struggle to do it that way so I try it the long way.
Title: Re: Calculus question....
Post by: Damo17 on December 13, 2008, 12:07:53 pm: Another one I'm stuck on.

Differentiate:
$y= x^2(1-6x)^{-3}$

This is what I have done so far:

$let u= x^{2}$

$\frac{du}{dx}= 2x$

$let v=(1-6x)^{-3}$

$\frac{dv}{dx}=18(1-6x)^{-4}$

$\frac{dy}{dx}=u\cdot \frac{dv}{dx}+v\cdot \frac{du}{dx}$

$= x^2\cdot18(1-6x)^{-4} + (1-6x)^{-3}\cdot 2x$

$=18x^2(1-6x)^{-4} + 2x(1-6x)^{-3}$

I think it is correct so far but I don't know where to go from here. Any help/explanation would be greatly appreciated.
Title: Re: Calculus question....
Post by: ed_saifa on December 13, 2008, 12:11:43 pm: Take out a common factor $(1-6x)^{-3}$
Title: Re: Calculus question....
Post by: Damo17 on December 13, 2008, 12:18:10 pm: The answer in the back of the book says:

$\frac{2x+6x^2}{(1-6x)^4}$

Which even with taking out the common factor I don't know how they get that.
Am I missing something really stupid?
Title: Re: Calculus question....
Post by: ed_saifa on December 13, 2008, 12:20:22 pm: Lowest common denominator?
Title: Re: Calculus question....
Post by: Damo17 on December 13, 2008, 12:38:51 pm: I'm still stuck, could someone show some working out. Please. :)
Title: Re: Calculus question....
Post by: shinny on December 13, 2008, 12:40:54 pm: I'm lazy with latex, but from the last step previously, take out a factor of $(1-6x)^{-4}$ , and then just change the power of -4 to make it go to the denominator. After that, a bit of simplifying should get you to your answer.
Title: Re: Calculus question....
Post by: Damo17 on December 13, 2008, 12:53:45 pm: Quote from: shinjitsuzx on December 13, 2008, 12:40:54 pm
I'm lazy with latex, but from the last step previously, take out a factor of $(1-6x)^{-4}$ , and then just change the power of -4 to make it go to the denominator. After that, a bit of simplifying should get you to your answer.

How can you take out $(1-6x)^{-4}$ when there aren't two of them?

I am sorry if I am getting a bit annoying but I still don't understand.
Title: Re: Calculus question....
Post by: dekoyl on December 13, 2008, 01:19:34 pm: Stuff LaTex :P It takes me 30 seconds to get it to computer + host.
Here's how I did it.
(http://i37.tinypic.com/2ev6xi1.png)
Title: Re: Calculus question....
Post by: Damo17 on December 13, 2008, 01:24:54 pm: ^
thanks sooo much. I get it now (finally). Thanks a million. :smitten: :smitten:
Title: Re: Calculus question....
Post by: Damo17 on December 16, 2008, 10:36:49 am: Another question:

Differentiate: $y= \frac{2x}{\sqrt{x+1}}$

This is what I have so far:

$let u= 2x$ $v=\sqrt{x+1}$
$\frac{du}{dx}=2$ $\frac{dv}{dx}= \frac{1}{2 \sqrt{x+1}}$

$\frac{dy}{dx}= \frac{2\sqrt{x+1} - 2x(x+1)^\frac{-1}{2}}{2(x+1)}$

Where do I go from here?
Title: Re: Calculus question....
Post by: unknown id on December 16, 2008, 10:41:59 am: Quote from: Damo17 on December 16, 2008, 10:36:49 am
Another question:

Differentiate: $y= \frac{2x}{\sqrt{x+1}}$

This is what I have so far:

$let u= 2x$ $v=\sqrt{x+1}$
$\frac{du}{dx}=2$ $\frac{dv}{dx}= \frac{1}{2}\sqrt{x+1}$

$\frac{dy}{dx}= \frac{2\sqrt{x+1} - 2x\sqrt{x+1}}{2(x+1)}$

Where do I go from here?

I just want to point out that $\frac{dv}{dx}= \frac{1}{{2}\sqrt{x+1}}$ , and not $\frac{dv}{dx}= \frac{1}{2}\sqrt{x+1}$

Then:

$\frac{dy}{dx}= \frac{2\sqrt{x+1} - \frac{x}{\sqrt{x+1}}}{x+1}$
Title: Re: Calculus question....
Post by: Damo17 on December 16, 2008, 10:43:23 am: Quote from: unknown id on December 16, 2008, 10:41:59 am
Quote from: Damo17 on December 16, 2008, 10:36:49 am
Another question:

Differentiate: $y= \frac{2x}{\sqrt{x+1}}$

This is what I have so far:

$let u= 2x$ $v=\sqrt{x+1}$
$\frac{du}{dx}=2$ $\frac{dv}{dx}= \frac{1}{2}\sqrt{x+1}$

$\frac{dy}{dx}= \frac{2\sqrt{x+1} - 2x\sqrt{x+1}}{2(x+1)}$

Where do I go from here?

I just want to point out that $\frac{dv}{dx}= \frac{1}{{2}\sqrt{x+1}}$ , and not $\frac{dv}{dx}= \frac{1}{2}\sqrt{x+1}$

Yeah I know, i'm having a little trouble with latex.
Title: Re: Calculus question....
Post by: cara.mel on December 16, 2008, 10:46:51 am: Quote from: Damo17 on December 16, 2008, 10:36:49 am
Another question:

Differentiate: $y= \frac{2x}{\sqrt{x+1}}$

This is what I have so far:

$let u= 2x$ $v=\sqrt{x+1}$
$\frac{du}{dx}=2$ $\frac{dv}{dx}= \frac{1}{2\sqrt{x+1}}$

$\frac{dy}{dx}= \frac{2\sqrt{x+1} - 2x\sqrt{x+1}}{2(x+1)}$

Where do I go from here?

Firstly, $\frac{dv}{dx}= \frac{1}{2\sqrt{x+1}}$ (possibly typo in latex :P)
Then
$\frac{dy}{dx} = \frac{2\sqrt{x+1}-\frac{x}{\sqrt{x+1}}}{x+1}$

Timsing both sides by $\sqrt{x+1}$
$\frac{dy}{dx} = \frac{2(x+1)-x}{(x+1)^{3/2}}$ = $\frac{x+2}{(x+1)^{3/2}}$
Title: Re: Calculus question....
Post by: Damo17 on December 16, 2008, 10:55:33 am: Quote from: caramel on December 16, 2008, 10:46:51 am
Quote from: Damo17 on December 16, 2008, 10:36:49 am
Another question:

Differentiate: $y= \frac{2x}{\sqrt{x+1}}$

This is what I have so far:

$let u= 2x$ $v=\sqrt{x+1}$
$\frac{du}{dx}=2$ $\frac{dv}{dx}= \frac{1}{2\sqrt{x+1}}$

$\frac{dy}{dx}= \frac{2\sqrt{x+1} - 2x\sqrt{x+1}}{2(x+1)}$

Where do I go from here?

Firstly, $\frac{dv}{dx}= \frac{1}{2\sqrt{x+1}}$ (possibly typo in latex :P)
Then
$\frac{dy}{dx} = \frac{2\sqrt{x+1}-\frac{x}{\sqrt{x+1}}}{x+1}$

Timsing both sides by $\sqrt{x+1}$
$\frac{dy}{dx} = \frac{2(x+1)-x}{(x+1)^{3/2}}$ = $\frac{x+2}{(x+1)^{3/2}}$

Thanks I understand where I went wrong. I should have canceled out the 2's in $\frac{dv}{dx}$ . Having 2 on the bottom for all ended up getting me confused.