Post by: Dragomistress on February 11, 2017, 08:07:47 am
I am rather confused as to how to set these questions out, any help would be greatly appreciated.
Thank you!
Post by: jamonwindeyer on February 11, 2017, 08:55:21 am
How do I do a and e?
I am rather confused as to how to set these questions out, any help would be greatly appreciated.
Thank you!
Hey mate! Lemme show you:
So that is the process, remember that \(f(x+h)\) is just whatever function we are talking about, with \((x+h\) instead of \(x\) wherever it appears. It will always work out that some terms up the top will cancel, then you'll be able to divide through by \(h\) as I did in the second last step ;D
E is the same but more complicated:
See if you can expand and simplify that top line and take it from there, you should get \(2x+3\), if it doesn't work out I'll finish it for you ;D
Post by: Dragomistress on February 11, 2017, 10:05:26 am
When am I supposed to use it?
Post by: Drewballs on February 11, 2017, 10:10:44 am
Post by: jamonwindeyer on February 11, 2017, 10:15:27 am
Just curious, on the Cambridge textbook examples, it uses a statement: "h (is not equal to) 0" before the final answer.
When am I supposed to use it?
The explanation for the \(h\neq0\) statement appearing where it does is that you've divided through by \(h\) from the previous step. You wouldn't be allowed to do this after evaluating the limit, because as soon as you substitute zero, it breaks (as Drew mentioned). You need to divide through by \(h\) first and the statement justifies that.
To be honest I think it is a little excessive. You don't need it, you'd get full marks in this sort of question without writing that - I've never seen it in a BOSTES sample solution ;D
Post by: Dragomistress on February 11, 2017, 02:04:01 pm
Post by: jamonwindeyer on February 11, 2017, 02:36:50 pm
May I have a sample of BOSTES answer to one of these questions. I can't seem to find one. Probably because I don't understand what I am doing :P
It doesn't pop up much in the HSC these days, here is a sample from Question 7(a)(i) 2009 (that is the Extension Paper) ;D I can't find any more on a quick search, it used to be more common, if you go to the NESA (new BOSTES) website and go through sample solutions from 2U exams (early 2000's late 90's) I think you'll find more there :)
But as an indicator, the working I gave for one of the questions above is roughly what you are aiming for :)
Post by: Dragomistress on February 11, 2017, 02:58:09 pm
Thanks a ton. I understand this concept now. 100 ATAR here I come :P
Post by: jamonwindeyer on February 11, 2017, 03:01:23 pm
Okay. Cool.
Thanks a ton. I understand this concept now. 100 ATAR here I come :P
No worries! Glad to hear it - This gave me a fair bit of trouble when I first learned it :)
Post by: Dragomistress on February 12, 2017, 11:15:27 am
According to the textbook:
If f(x)=g(x)+h(x), then
f'(x)=g'(x)+h'(x)
Question
Differentiate: f(x)=4x^2-3x+2
Why is this wrong?
f'(x)= 4(2x)-3(x^0)+2
f'(x)=8x-3+2
f'(x)=8x-1
But the actual answer is 8x-3. Does the two get ignored or something?
Also, how do I make that beautiful looking type font for math?
:D
Post by: RuiAce on February 12, 2017, 11:20:29 am
Oh wow... I'm stuck again :/The derivative of 2 is 0. Everything else was good.
According to the textbook:
If f(x)=g(x)+h(x), then
f'(x)=g'(x)+h'(x)
Question
Differentiate: f(x)=4x^2-3x+2
Why is this wrong?
f'(x)= 4(2x)-3(x^0)+2
f'(x)=8x-3+2
f'(x)=8x-1
But the actual answer is 8x-3. Does the two get ignored or something?
Also, how do I make that beautiful looking type font for math?
:D
This is my LaTeX guide. There's more of these floating around the internet.
Post by: Dragomistress on February 13, 2017, 08:39:14 pm
Thanks
Post by: kiwiberry on February 13, 2017, 08:42:57 pm
I would like someone to show me the working out as the textbook doesn't show the substitution for dy/dx and I am confused on how this substitution works. Please show through 1a.
Thanks
Post by: Dragomistress on February 14, 2017, 08:59:55 pm
May someone show me the quickest way to do this question (just so that I can verify and potentially upgrade my working).
[This is a chain rule derivative question]
Post by: Shadowxo on February 14, 2017, 10:09:21 pm
Cool, thanks for the replies :D Best forum of my life.
May someone show me the quickest way to do this question (just so that I can verify and potentially upgrade my working).
[This is a chain rule derivative question]
Great answer by M909! :)
Just a little extra:
dy/dx = dy/du * du/dx.
Let's say u = px + q
y = u8
dy/du = 8u7
du/dx = p
dy/dx = 8u7*p = 8p(px+q)7
So if you've got anything inside a bracket to a power that you need to derive, you derive the bracket section, treating what's inside as u then multiplying by the derivative of the bracket
eg 2(3x2+6x+12)5
derivative is 2*5(3x2+6x+12)4*(6x+6)
Hope this helps :)
Post by: Dragomistress on February 15, 2017, 04:07:51 pm
Or is the same as dy/dx?
Post by: RuiAce on February 15, 2017, 04:13:07 pm
So what does the d/dx mean?
Or is the same as dy/dx?
Post by: Dragomistress on February 15, 2017, 04:37:07 pm
For example solve the derivative of: (3x+4)^3
d/dx (3x+4)^3 = 9x(3x+4)^2
Instead of writing
Let y = (3x+4)^3
dy/dx = (3x+4)^3
= 9x(3x+4)^2
Post by: jakesilove on February 15, 2017, 04:51:54 pm
So I use the d/dy when there is no designated letter for a value for convenience?
For example solve the derivative of: (3x+4)^3
d/dx (3x+4)^3 = 9x(3x+4)^2
Instead of writing
Let y = (3x+4)^3
dy/dx = (3x+4)^3
= 9x(3x+4)^2
Yep that's perfect!
Post by: Dragomistress on February 15, 2017, 06:14:53 pm
How do I do this using the quick formula?
Post by: jakesilove on February 15, 2017, 06:35:54 pm
Now I just feel like I am abusing the forum for questions :/ but I just truly don't understand.
How do I do this using the quick formula?
The quick formula you are looking for here is the chain rule. It goes like this.
So, here's what I want you to do. Write each question out in your work book. Identify f(x), and write it out. For instance, in a), f(x)=3x+7. Then, find f'(x) (ie. the first derivative of f(x)). Write out f'(x). For instance, in a), f'(x)=3. Then, plug the values you've found into the formula above! That's all there is to it.
Post by: jamonwindeyer on February 16, 2017, 12:53:24 am
Now I just feel like I am abusing the forum for questions :/
You could never! If you are benefitting from the help then our time is well spent - It's what we are here for ;D