Help with this C4 question

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 $x$. The LHS would require the use of the product rule.

how do I do the first step?

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?

Im so stuck lol

Im so stuck lol

Im so stuck lol

This should've been covered in lessons, when you differentiate a variable $y$ with respect to another $x$, you just end up with $\dfrac{dy}{dx}$.

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.

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