Question

find the derivative of (x-y)^2=y^2-xy

Answer 1

something about implicit differentiation i think

Answer 2

implicit rules are the same as explicit rules

Answer 3

\[D[(x-y)^2=y^2-xy]\]
\[D[(x-y)^2]=D[y^2]-D[xy]\]
\[2(x-y)*D[x-y]=2yy'-(D[x]y+xD[y])\]
\[2(x-y)*(D[x]-D[y])=2yy'-(x'y+xy')\]
\[2(x-y)(x'-y')=2yy'-(x'y+xy')\]

Answer 4

when you determine what this is "with respect to" then you can determine what to do with the ' parts

Answer 5

it says to find dy/dx

Answer 6

dy/dx = y'

dx/dx = x' = 1

Answer 7

\[2(x-y)(x'-y')=2yy'-(x'y+xy')\]
\[2(x-y)(1-y')=2yy'-(y+xy')\]
do the algebra to solve for y'

Answer 8

the key tho, is to realize the implicits are no different than explicits; the rules dont change, just the manner in which your used to seeing them changes - basically because they teach this stuff in a method manner instead of a comprehesive manner

Answer 9

i got the formula and the rules right, but when i solve for y' i keep getting the wrong answer

Answer 10

yeah, algebra can be a pain :)

can you show me your steps?

Answer 11

(2x-2y)(1-y')=2yy'-y-xy'
2x-2xy'-2y+2yy'=2yy'-y-xy'
xy'=y+2x-2y
xy'=2x-y
y'=(2x-y)/x
oh nevermind i got it :)

Answer 12

thank you so much

Answer 13

yay!! :)