Question

How do I do implicit differentiation on the equation

x^2 -4xy +y^2 = 4 ?

Answer 1

treat y as a function of x
so the derivative with respect to x, of y^2 is 2y.dy/dx
use the product rule to differentiate 4xy

find the derivative term by term then make dy/dx the subject

Answer 2

does this help?

Answer 3

I get Y -4dy/dx for the -4xy. is this correct?

Answer 4

no its -(4x * dy/dx + 4y)

= -4x dy/dx - 4y

Answer 5

so we have
2x - 4x.dy/dx - 4y + 2y.dy/dx = 0

now solve for dy/dx

Answer 6

I get Y Prime = (x+3y)/ -2x

I know this is not the answer

Answer 7

I should be getting (2y-x)/(y-2x)

Answer 8

But I'm not.

Answer 9

because I know the answer, but that does me no good if I don't understand how to get the answer. The part that is tripping me up the most is that -4xy. we use y' instead of dy/dx in our class.

Answer 10

2x - 4x.dy/dx - 4y + 2y.dy/dx = 0
2y.dy/dx - 4x dydx = 4y - 2x
dy/dx = (4y - 2x) / ( 2y - 4x)

dividing top and bottm by 2

dy/dx = (2y - x)/(y - 2x)

Answer 11

can you explain how you get to your answer

Answer 12

derivative of -4xy
= derivative of - (4xy)

= - ( 4x.y' + y* 4)
= -4xy' - 4y

Answer 13

Product Rule:

f'(xy) = f(x)f'(y) + f(y)f'(x)

Answer 14

the derivative of y^2 is found using the chain rule

Answer 15

it's mainly the algebra that is messing me up . being able to get 2x - 4xy'-4y=2yy' into the form (2y-x)/ (y-2x)

Answer 16

I am having trouble getting the Y' by itself on one side.

Answer 17

2y.dy/dx - 4x dydx = 4y - 2x
= 2y y' - 4x y' = 4y - 2x
factor the left side
(y' is common to both terms):-
y' (2y - 4x) = 4y - 2x
Now divide both sides by 2y-4x:-
y' = (4y - 2x)/(2y - 4x)
dividing by 2 gives
y' = (2y - x)/(y - 2x)

Answer 18

* dividing top an d bottom of fraction by 2

Answer 19

alright. that clears it up. thank you

Answer 20

yw