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# C4 differentiation Tweet

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1. C4 differentiation
hi when we are asked to find the equation of line and if we get like 6xy we need to use the product rule and get 6y+6x but how do we know to put dy/dx like do we put it with 6y or 6x?
2. Re: C4 differentiation
(Original post by otrivine)
hi when we are asked to find the equation of line and if we get like 6xy we need to use the product rule and get 6y+6x but how do we know to put dy/dx like do we put it with 6y or 6x?
you do it like; 6x is u, and y is v. u'v + v'u = 6y + 1dy/dx ..
3. Re: C4 differentiation
(Original post by otrivine)
hi when we are asked to find the equation of line and if we get like 6xy we need to use the product rule and get 6y+6x but how do we know to put dy/dx like do we put it with 6y or 6x?

Have you learnt about implicit differentiation
4. Re: C4 differentiation
(Original post by demdil)
you do it like; 6x is u, and y is v. u'v + v'u = 6y + 1dy/dx ..
You seem to have lost a 6x
5. Re: C4 differentiation
(Original post by TenOfThem)
Have you learnt about implicit differentiation
yes my method is correct but i put 6y.dy/dx but mark scheme but 6x.dy/dx why and how?
6. Re: C4 differentiation
(Original post by TenOfThem)
You seem to have lost a 6x
oh yes thannk you then it's 6y + 6x(dy/dx) ..
7. Re: C4 differentiation
(Original post by otrivine)
yes my method is correct but i put 6y.dy/dx but mark scheme but 6x.dy/dx why and how?
6xy;
6x = u
y = v
u' = 6
v' = dy/dx

u'v + v'u = 6y + dy/dx times 6x
= 6y + 6x(dy/dx)
8. Re: C4 differentiation
(Original post by demdil)
6xy;
6x = u
y = v
u' = 6
v' = dy/dx

u'v + v'u = 6y + dy/dx times 6x
= 6y + 6x(dy/dx)
thats the thing it confuses i thought we usually put dy/dx always with Ys not with Xs?
9. Re: C4 differentiation
(Original post by otrivine)
yes my method is correct but i put 6y.dy/dx but mark scheme but 6x.dy/dx why and how?

You have said that

and it isn't it is 1
10. Re: C4 differentiation
(Original post by otrivine)
thats the thing it confuses i thought we usually put dy/dx always with Ys not with Xs?
Do you understand implicit differentiation and the chain rule
11. Re: C4 differentiation
(Original post by otrivine)
thats the thing it confuses i thought we usually put dy/dx always with Ys not with Xs?
i dont know why but i think thats a rule. so in this type of question put a dy/dx every time you differentiate y
12. Re: C4 differentiation
(Original post by demdil)
i dont know why but i think thats a rule. so in this type of question put a dy/dx every time you differentiate y
yes i used the chain rule and got 6y+6x correct then how would i know where to put dy/dx this is my question?
13. Re: C4 differentiation
(Original post by TenOfThem)
Do you understand implicit differentiation and the chain rule
yes i used the chain rule and got 6y+6x correct then how would i know where to put dy/dx this is my question?
14. Re: C4 differentiation
(Original post by otrivine)
yes i used the chain rule and got 6y+6x correct then how would i know where to put dy/dx this is my question?
v = y
v' = dy/dx

u'v + v'u = 6y + dy/dx times 6x(as u=6x)
= 6y + 6x.dy/dx

so you'll put it when you differentiate y ..
15. Re: C4 differentiation
(Original post by otrivine)
yes i used the chain rule and got 6y+6x correct then how would i know where to put dy/dx this is my question?
The chain rule tells you that

When you try to do you have to use the chain rule so
16. Re: C4 differentiation
(Original post by demdil)
v = y
v' = dy/dx

u'v + v'u = 6y + dy/dx times 6x(as u=6x)
= 6y + 6x.dy/dx

so you'll put it when you differentiate y ..
ohhhhh yes i got it thanks for your help ++reps
17. Re: C4 differentiation
(Original post by TenOfThem)
The chain rule tells you that

When you try to do you have to use the chain rule so
yeesss i got it now thanks tenofthem you were helpful ++repsssss