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

Expand to get xy+3y-4x-12=x^2+y^2Rearrange 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 =ydu = 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

Answered by Ravi R. Maths tutor

3841 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Given that y= 1/ (6x-3)^0.5 find the value of dy/dx at (2;1/3)


If the function f is defined as f= 1-2x^3 find the inverse f^-1


Can you help me understand how Arithmetic sequences work?


Differentiate a^x


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences