Question

find dy/dx of the following
1) x^2+y^2-a^2=0
2) x^2/a^2+y^2/b^2=1
3) x^3-4xy^2-2y^2-5y^3=0
plz help...:(

Answer 1

.1. 2x + 2y.dy/dx = 0
dy/dx = -x/y

2. 2x/a^2 +( 2y/b^2 ). dy/dx = 0
dy/dx = - 2x/a^2 / 2y/b^2
= -2x * b^2
--- ---
a^2 2y

= - x b^2
----
y a^2

Answer 2

3. 3x^2 - (4x 2y. dy/dx +4y^2) - 4y .dy/dx -15y^2 . dy/dx = 0
3x^2 - 8xy. dy/dx - 4y^2 - 4y .dy/dx -15y^2 . dy/dx = 0
8xy. dy/dx + 4y .dy/dx + 15y^2 . dy/dx = 3x^2
dy/dx = 3x^2 / (8xy + 4 + 15y^2)

Answer 3

which bit are u talking about?

Answer 4

???

Answer 5

oh i understand now thanx...:)

Answer 6

y w