Question

derivative:
(x+y)^2 = x^2 + y^2

Answer 1

partials or what?

Answer 2

i dont know it just says find y' in terms of x and y using implicit

Answer 3

xy'+y=0

Answer 4

then im pretty sure it has to do with "respect to x" jargon. implicit it is

The key to remember is that there are derived bits that pop out when we take any derivative, implicit or explicit. But with explicit they tend to throw them away before you realized what they were; but also kept them in so that you could solve for the problem...

for instance; y' IS a derived bit that they keep in. x' is a derived but that they just toss out (because it has a value of 1).

But for implicits, lets keep all our derived bits till the end shall we.

(x+y)^2 = x^2 + y^2 ; derive each part seperately

[(x+y)^2]' = [x^2]' + [y^2]' ; and use all the rules of
derivatives you know of

2(x+y) [x+y]' = 2x x' + 2y y' ; divide out the 2
/2 /2 /2
-------------------------
(x+y) [x+y]' = x' + y' ; solve [x+y]'

(x+y)(x'+y') = x'+y' ; expand and solve for y'

x+y
x'+y'
------
xx' + yx'
y'x + y'y
------------
xx'+yx'+y'x+y'y = x'+y'

y'x + y'y - y' = x' - xx' -yx' ; now that y's are all to one side...

y'(x+y-1) = x'(1 -x -y) ; we might as well factor out an x'
as well, but solve for y' now

x'(1-x-y)
y' = --------- ; the x' is cleaner notation for dx/dx;
(x+y-1) which equals 1 sooo

1-x-y
y' = ------- ...as long as i kept the math right
x+y-1