dy/dx ????

Question

if (x-y)^2=y^2-xy then dy/dx=
a)(2x-y)/(2y-x)
b)(2x-y)/2x
c)(2x-y)/x
d)(2x+3y)/x
e)undefined

Answer 1

Use the chain rule for (x-y)^2
use the product rule for xy

2(x-y) *(1-dy/dx) = 2y*dy/dx - (y + x*dy/dx)
(2x-2y)(1-dy/dx) = 2y*dy/dx - y - x*dy/dx
2x - 2x*dy/dx -2y +2y*dy/dx = 2y*dy/dx - y - x*dy/dx
2x-2y+y= 2y*dy/dx - x*dy/dx +2x*dy/dx - 2y*dy/dx
2x-y = x * dy/dx
(2x-y)/x = dy/dx

[Answer: C]

Answer 2

c.
1st method:
x^2-2xy+y^2=y^2-xy
x^2-2xy=-xy
xy=x^2
y=x
dy/dx=1
(2x-y)/x=(2x-x)/x=x/x=1

(2x-y)/x undefined for x=0

2nd explanation:
x^2-2xy+y^2=y^2-xy
Differentiating, we get:
2x-2y-2xy'+2yy'=2yy'-y-xy'
2yy' can be subtracted from both sides; putting all terms involving y' on the right side, we get:
2x-y=xy'
Dividing both sides by x gives us:
y'=(2x-y)/x, or dy/dx=(2x-y)/x

Answer 3

(x-y)^2 = y^2 - xy

simplify

x^2 + y^2 - 2xy = y^2 - xy

x^2 - xy = 0

differentiating implicitly

2x - xy' - y = 0

xy' = 2x - y

y' = (2x - y) / x

Answer 4

(x-y)^2=y^2-xy
differentiate both sides with respect tox
2(x-y)(1-y')=2yy'-xy'-y
2x-2y-2xy'+2yy' = 2yy' -xy'-y
2x-3y=3xy'
y' = (2x-3y)/3x

Answer 5

(x-y)² = y²-xy
2(x-y)(1-y') = 2yy' - y - xy'
2(x - xy' - y + yy') = 2yy' - y - xy'
2x - 2xy' - 2y + 2yy' = 2yy' - y - xy'
-xy' = y -2x
y' = (2x-y)/x
dy/dx = (2x-y)/x

(c)