You are Here: Home >< Maths

# Help with this C4 question watch

1. the equation of a curve C is (x+3)(y-4)=x^2+y^2
find dy/dx in terms of x and y

Thank you!
2. (Original post by Astana)
the equation of a curve C is (x+3)(y-4)=x^2+y^2
find dy/dx in terms of x and y

Thank you!
What have you tried?

You can begin by differentiating each side w.r.t . The LHS would require the use of the product rule.
3. using chain rule ;
(d (y^n))/dx = ny^(n-1)dy/dx
And product rule;
xdy/dx + y.

I hope u can take it on from here
let me know if this was helpful
4. (Original post by RDKGames)
What have you tried?

You can begin by differentiating each side w.r.t . The LHS would require the use of the product rule.
how do I do the first step?
5. (Original post by Astana)
how do I do the first step?
Differentiate the LHS w.r.t

So, by the product rule,

What do you get?
6. (Original post by brainmaster)
using chain rule ;
(d (y^n))/dx = ny^(n-1)dy/dx
And product rule;
xdy/dx + y.

I hope u can take it on from here
let me know if this was helpful
Thank you, this is very helpful. How do you differentiate the y-4 though?
7. Im so stuck lol
8. (Original post by Astana)
Thank you, this is very helpful. How do you differentiate the y-4 though?
we know that d (y^n)/dx = ny^(n-1)dy/dx
so lets differentiate y-4 with respect to x. this means we will differentiate each term separately;
we can write 4 as 4y^0 so we differentiating;
y - 4y^0
so dy/dx = 1*y^(1-1) *(dy/dx) + (-4*0*y^(0-1)*(dy/dx)
if we simplify we get;
1*y^0*dy/dx - 4*0*y^-1*dy/dx
but y^0 = 1 and the other part has been multiplied by 0 henxe the whole thing will be 0
so we get;
1*1*dy/dx - 0
which is further simplified we get;
dy/dx.
so this means that;
d (y-4)/dx = dy/dx

hope u understood this
9. (Original post by Astana)
Im so stuck lol
This should've been covered in lessons, when you differentiate a variable with respect to another , you just end up with .

This is what you've been doing all the time in C1 without thinking about it. You differentiated both sides and ended up with dy/dx on one of them.
10. (Original post by Astana)
Im so stuck lol
it's okay you can ask anything you don't understand coz I'm sure once you get the idea you will be able to do other questions
11. Expand to get xy+3y-4x-12=x^2+y^2
Rearrange to xy+3y-4x-12-x^2-y^2

Then you have to differentiate implicitly:

use product rule on xy to get:
u = x v =y
du = 1 dv = dy/dx

so xy differentiates to xdy/dx + y

Then when you differentiate 3y you get:
3dy/dx

-4x-12-x^2 differentiates to -4-2x

-y^2 differentiates to -2ydy/dx

so when you combine that all you get:

xdy/dx+y+3dy/dx-4-2x-2ydy/dx=0

rearrange to get:

xdy/dx+3dy/dx-2ydy/dx=-y+4+2x

factorise out dy/dx:

dy/dx(x+3-2y)=-y+4+2x

Then finally divide across:

dy/dx = -y+4+2x/x+3-2y
12. (Original post by Astana)
Im so stuck lol
let me know if you are now able to do this question and once u do it, try 4F mixed exercise C4 question number 16 since it needs the same idea good luck
13. (Original post by RDKGames)
This should've been covered in lessons, when you differentiate a variable with respect to another , you just end up with .

This is what you've been doing all the time in C1 without thinking about it. You differentiated both sides and ended up with dy/dx on one of them.
not in my lessons lol, thank you so much!!
14. (Original post by SplendidHoney)
Expand to get xy+3y-4x-12=x^2+y^2
Rearrange to xy+3y-4x-12-x^2-y^2

Then you have to differentiate implicitly:

use product rule on xy to get:
u = x v =y
du = 1 dv = dy/dx

so xy differentiates to xdy/dx + y

Then when you differentiate 3y you get:
3dy/dx

-4x-12-x^2 differentiates to -4-2x

-y^2 differentiates to -2ydy/dx

so when you combine that all you get:

xdy/dx+y+3dy/dx-4-2x-2ydy/dx=0

rearrange to get:

xdy/dx+3dy/dx-2ydy/dx=-y+4+2x

factorise out dy/dx:

dy/dx(x+3-2y)=-y+4+2x

Then finally divide across:

dy/dx = -y+4+2x/x+3-2y
Omg, thank you so much, this was very easy to follow - will use this method in the future

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 12, 2018
The home of Results and Clearing

### 955

people online now

### 1,567,000

students helped last year
Today on TSR

### University open days

1. Keele University
Sun, 19 Aug '18
2. University of Melbourne
Sun, 19 Aug '18
3. Sheffield Hallam University
Tue, 21 Aug '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams