Mathematics: Post your doubts here!

[fleurisabelle]

badly stuck in Chain Rule...somebody help please
I know that dy/dx is gradient with respect to x but why the hell do they start using du/dx and d/dy, all of a sudden and what exactly does it mean?

 

[Anum96]

badly stuck in Chain Rule...somebody help please
I know that dy/dx is gradient with respect to x but why the hell do they start using du/dx and d/dy, all of a sudden and what exactly does it mean?

dy/dx is not just the gradient. It means differentiation of y with respect to x. Du/dx means differentiation of u with respect to x. In chain rule you have a fraction.
lets say
2x - 1
--------
3x

You usually assume numerator to be u and denominator to be v ( Its just the process of assigning variables )
and differentiate them separately.
u is 2x - 1
differentiating it will give you du/dx (its also written as u')as 2
v is 3x
differentiating it will give you dv/dx ( or v' ) as 3
Now you have these two answers just plug them in the chain rule formula which is given in your formula booklet.
( vu' - uv' )/ v^2
OR
v(du/dx) - u(dv/dx) / v^2

[(3x)(2) - (2x - 1)(3)] / (3x)^2

 

[fleurisabelle]

dy/dx is not just the gradient. It means differentiation of y with respect to x. Du/dx means differentiation of u with respect to x. In chain rule you have a fraction.
lets say
2x - 1
--------
3x

You usually assume numerator to be u and denominator to be v ( Its just the process of assigning variables )
and differentiate them separately.
u is 2x - 1
differentiating it will give you du/dx (its also written as u')as 2
v is 3x
differentiating it will give you dv/dx ( or v' ) as 3
Now you have these two answers just plug them in the chain rule formula which is given in your formula booklet.
( vu' - uv' )/ v^2
OR
v(du/dx) - u(dv/dx) / v^2

[(3x)(2) - (2x - 1)(3)] / (3x)^2

thanks a lot...that did help me, pretty much explained for the Chain Rule but the thing is that when I reach differentiating implicit functions, that's were I'm stuck, they are actually now differentiating with respect to y, how do I do that?

 

[Anum96]

thanks a lot...that did help me, pretty much explained for the Chain Rule but the thing is that when I reach differentiating implicit functions, that's were I'm stuck, they are actually now differentiating with respect to y, how do I do that?

Just remember one rule for implicit function. Whenever you differentiate y in any implicit function you have to write 'dy/dx' along with it. You can easily separate the like terms during the rest of the working.
For e.g. You have 3(x^2) - (x^2)(y^3)
Your differentiation will be
So just differentiate and remember that rule
you will get 6x - [2xy^3 plus (x^2)(3y^2 (dy/dx))]
6x - 2xy^3 - (x^2)(3y^2 (dy/dx))

I dont have a proper question here but the concept is same. Youre usually asked to find dy/dx
So you just combine the like terms and make dy/dx the subject

 

[fleurisabelle]

Just remember one rule for implicit function. Whenever you differentiate y in any implicit function you have to write 'dy/dx' along with it. You can easily separate the like terms during the rest of the working.
For e.g. You have 3(x^2) - (x^2)(y^3)
Your differentiation will be
So just differentiate and remember that rule
you will get 6x - [2xy^3 plus 3y^2 (dy/dx)]
6x - 2xy^3 - 3y^2 (dy/dx)

I dont have a proper question here but the concept is same. Youre usually asked to find dy/dx
So you just combine the like terms and make dy/dx the subject

yesss yess yes ....I'm kinda getting the hold of it! but wait, (sorry for bothering you again and again but just the last thing) what is d/dy and d/dx then, they've used it so much (and in this Key Point in the screenshot)

 

[fleurisabelle]

Just remember one rule for implicit function. Whenever you differentiate y in any implicit function you have to write 'dy/dx' along with it. You can easily separate the like terms during the rest of the working.
For e.g. You have 3(x^2) - (x^2)(y^3)
Your differentiation will be
So just differentiate and remember that rule
you will get 6x - [2xy^3 plus (x^2)(3y^2 (dy/dx))]
6x - 2xy^3 - (x^2)(3y^2 (dy/dx))

I dont have a proper question here but the concept is same. Youre usually asked to find dy/dx
So you just combine the like terms and make dy/dx the subject

OMG! dunno what the hell happened to the screenshot

 

[Anum96]

yesss yess yes ....I'm kinda getting the hold of it! but wait, (sorry for bothering you again and again but just the last thing) what is d/dy and d/dx then, they've used it so much (and in this Key Point in the screenshot) View attachment 58137 View attachment 58137 View attachment 58137 View attachment 58137 View attachment 58137 View attachment 58137 View attachment 58137

Its written
(d/dx)(y^2)
Its the same as dy^2/dx Which means differentiation of y^2 with respect to x
And (d/dx)(f(x)) Is exactly the same as df(x)/dx which means differentiation of f(x) with respect to x
I hope you get it. Should you require any more detail, feel free to ask me

 

[Anum96]

OMG! dunno what the hell happened to the screenshot

Hahahaha! And I'm dumb enough to go through each one of them

 

[fleurisabelle]

Its written
(d/dx)(y^2)
Its the same as dy^2/dx Which means differentiation of y^2 with respect to x
And (d/dx)(f(x)) Is exactly the same as df(x)/dx which means differentiation of f(x) with respect to x
I hope you get it. Should you require any more detail, feel free to ask me

YEEES...I totally got it! Thank you so much!!!!!

 

[Anum96]

YEEES...I totally got it! Thank you so much!!!!!

Pleasure

 

[sj0007]

In part (iii), when they say 'B moves upwards', they mean rebound?
Can anyone draw the graph?
Thnx

 

[Anum96]

View attachment 58144
In part (iii), when they say 'B moves upwards', they mean rebound?
Can anyone draw the graph?
Thnx

There will be 2 lines for when the string is taut. First the graph will show a line with positive slope of 2.5 (your ans to part i)
Then the same line will come to rest after 1s ( your ans to part ii ) and will further move down
And for when the string is slack the second line segment will have a negative slope. (represented by A)
How can I attach a file here?

 

[Anum96]

AHA! There u go!

 

[Rizwan Javed]

View attachment 58145
AHA! There u go!

Wow nice sketching

 

[Anum96]

Wow nice sketching

Well, Thank you!
It was a paint file and I just couldn't find a way to attach it. MS Word helped

 

[Rizwan Javed]

Well, Thank you!
It was a paint file and I just couldn't find a way to attach it. MS Word helped

oh great

 

[Anum96]

oh great

 

[sj0007]

There will be 2 lines for when the string is taut. First the graph will show a line with positive slope of 2.5 (your ans to part i)
Then the same line will come to rest after 1s ( your ans to part ii ) and will further move down
And for when the string is slack the second line segment will have a negative slope. (represented by A)
How can I attach a file here?

View attachment 58145
AHA! There u go!

Lol........ Thank you soooo much!

 

[Anum96]

Lol........ Thank you soooo much!

Pleasuree!

 

[The Sarcastic Retard]

http://papers.gceguide.com/A Levels/Mathematics (9709)/9709_s15_qp_22.pdf
Q1)(ii) I got least value as -21.6 and greastest as 21.6 so how n can be 43?
Q3(i) why tan theta cant be negative 4?
Q4(i) I got dy/dx = e^x - 8e^-2x. Now how to solve this when dy/dx = 0?
I did like this,
e^x(1 - 8e^-3x) = 0
ln (1/8) = -3x.
Then is it like, x = ln(1/8) / -3 ?
Q5(ii)(iii) No idea.
Thanks.
Anum96 See if u can help me. ^_^