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# AQA C3 What type of differentiation topic would this type of question be? watch

1. Hi, I did a past paper the AQA C3 June 2012 one and I did everything fine apart from when I got to the final question and I didn't have a clue what to do.

The questions which I didn't understand are c and all of d.
Is there a specific name for this type of differentiation with stuff on the bottom, and differentiating tan^-1(ax-b)
as well as differentiating to find the second derivative when it is a fraction like the dy/dx in C.

If there isn't a name for the topic does anyone know where I can find some additional resources with more questions which I can practise on? I don't know how to search for these types of questions as its the more advanced differentiation of C3 which I can't do.
2. (Original post by Sayless)
Hi, I did a past paper the AQA C3 June 2012 one and I did everything fine apart from when I got to the final question and I didn't have a clue what to do.

The questions which I didn't understand are c and all of d.
Is there a specific name for this type of differentiation with stuff on the bottom, and differentiating tan^-1(ax-b)
as well as differentiating to find the second derivative when it is a fraction like the dy/dx in C.

If there isn't a name for the topic does anyone know where I can find some additional resources with more questions which I can practise on? I don't know how to search for these types of questions as its the more advanced differentiation of C3 which I can't do.
this is prelude of implicit differentiation which will eventually cover in your board in C4

In C3 you laern that

if x = f(y), then dy/dx = 1 / dx/dy
3. (Original post by TeeEm)
this is prelude of implicit differentiation which will eventually cover in your board in C4

In C3 you laern that

if x = f(y), then dy/dx = 1 / dx/dy
So these questions are both 1/dx/dy questions, still for d(ii) I wouldn't have a clue how to differentiate something like that i've never seen it before
4. (Original post by Sayless)
So these questions are both 1/dx/dy questions, still for d(ii) I wouldn't have a clue how to differentiate something like that i've never seen it before
For d)ii)
You simply need to differentiate:

As for the derivative of , you should calculate the derivative at least once using the method TeeEm suggested, but from then on, I would consider it a quotable result, and I imagine it's in your formula book.
5. (Original post by joostan)
For d)ii)
You simply need to differentiate:

As for the derivative of , you should calculate the derivative at least once using the method TeeEm suggested, but from then on, I would consider it a quotable result, and I imagine it's in your formula book.
Yeah, I have to differentiate that, but I don't know how to, I can differentiate using chain rule, product rule, quotient rule, but for stuff with stuff squared underneath in a fraction and another fraction I haven't learnt how to differentiate that, do you have a guide or something I could use.
I can't do TeeEm's method on y=tan^-1(x-1)
I ended up with dy/dx = 1/sec^2 y
when I use it from the formula booklet I get the correct answer though
6. For part di) I got x=2 and x=1 for part dii) do I need to use quotient rule
7. (Original post by Sayless)
Yeah, I have to differentiate that, but I don't know how to, I can differentiate using chain rule, product rule, quotient rule, but for stuff with stuff squared underneath in a fraction and another fraction I haven't learnt how to differentiate that, do you have a guide or something I could use.
I can't do TeeEm's method on y=tan^-1(x-1)
I ended up with dy/dx = 1/sec^2 y
when I use it from the formula booklet I get the correct answer though
Yes and you found that sec^2y is x^2-2x+2 so dy/dx=1/x^2-2x+2
8. (Original post by Sayless)
Yeah, I have to differentiate that, but I don't know how to, I can differentiate using chain rule, product rule, quotient rule, but for stuff with stuff squared underneath in a fraction and another fraction I haven't learnt how to differentiate that, do you have a guide or something I could use.
I can't do TeeEm's method on y=tan^-1(x-1)
I ended up with dy/dx = 1/sec^2 y
when I use it from the formula booklet I get the correct answer though
Look in your formula booklet: it should say that - so in your case, just replace the with .
9. (Original post by Hassan55)
Yes and you found that sec^2y is x^2-2x+2 so dy/dx=1/x^2-2x+2
Ah right yeah I get that part now thanks

(Original post by Zacken)
Look in your formula booklet: it should say that - so in your case, just replace the with .
I managed to do that now, stuck with d(ii) only now
10. Part dii) dy/dx=(1/x^2-2x+2) - (1/x) so differentiating first fraction using quotient rule u=1 v=x^2-2x+2 and second fraction would just be -lnx
11. (Original post by Sayless)
Ah right yeah I get that part now thanks

I managed to do that now, stuck with d(ii) only now
Well, you need to evalute then once you've got that, differentiate whatever you have again - what do you get?
12. (Original post by Hassan55)
For part di) I got x=2 and x=1 for part dii) do I need to use quotient rule
You don't HAVE to use the quotient rule, but you may as well.
13. (Original post by joostan)
You don't HAVE to use the quotient rule, but you may as well.
What other methods are there in order to differentiate a quotient?
14. (Original post by Zacken)
Well, you need to evalute then once you've got that, differentiate whatever you have again - what do you get?
I'm trying to do the quotient rule for it, I get
u=1
v= x^2 -2x + 2
du/dx = 0
dv/dx = 2x-2

(x^2-2x+2)(0) - (1)(2x-2) / (x^2-2x+2)^2

= -2x+2 / (x^2-2x+2)^2

then for the 1/x i get x^-2

so overall its
-2x+2/(x^2-2x+2)^2 + x^-2

is that right
15. (Original post by Sayless)
I'm trying to do the quotient rule for it, I get
u=1
v= x^2 -2x + 2
du/dx = 0
dv/dx = 2x-2

(x^2-2x+2)(0) - (1)(2x-2) / (x^2-2x+2)^2

= -2x+2 / (x^2-2x+2)^2

then for the 1/x i get x^-2

so overall its
-2x+2/(x^2-2x+2)^2 + x^-2

is that right
Almost d/dx(1/x) = -x^(-2) not +
16. (Original post by Hassan55)
What other methods are there in order to differentiate a quotient?
This is simply the chain rule with
17. (Original post by Zacken)
This is simply the chain rule with
Oh yeah of course, thanks
18. (Original post by Zacken)
Almost d/dx(1/x) = -x^(-2) not +
Thanks, i got it now
19. (Original post by Sayless)
Thanks, i got it now
Great!

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