Question

differentiating (x-y)^2=x+y-1

Answer 1

wrt x yes?

Answer 2

yes

Answer 3

so you start with
\[(x-y)^2=x+y+1\]
\[2(x-y)(1+y')=1+y'\]

Answer 4

is that much ok?

Answer 5

in implicit differentiation you are thinking that
\[y=f(x)\] so
\[y'=f'(x)\]

Answer 6

I don't know how to get to the next step:
2 (x-y) - 2 (x-y) y' = 1 + y'
are you using the chain rule or something to get to it?

Answer 7

oh if the first step is ok the we are on the way

Answer 8

now it is just annoying algebra to solve for y'

Answer 9

er. i mean how to get to the step after the one you posted and to the step that i posted.

Answer 10

i guess you have to multiply out on the left, collect terms, put everything with y' on one side and everything else on the other, factor and divide

Answer 11

that is the answer that the solution manual gives, but it is lacking a coherent explanation

Answer 12

the fist step is the only "calculus " step. it is algebra from here on in

Answer 13

we can work it out step by step if you like

Answer 14

yes please :)

Answer 15

ok but first is this ok?
\[2(x-y)(1+y')=1+y'\]?

Answer 16

yes

Answer 17

if not we should work that out first

Answer 18

that is actually where i got stuck on

Answer 19

wait you got stuck on first step? then lets do it slowly

Answer 20

first of all right hand side is just
\[x+y+1\] and we are thinking that
\[y=f(x)\] even though we don't know what
\[f(x)\] is i.e. we have not "solved" for it

Answer 21

that is why it is called "implicit differentiation because we are thinking of y "implicitly" as a function of x

Answer 22

so if we write the right hand side as \[x+f(x)+1\] and take derivatives we get
\[1+f'(x)\] because the derivative of x is 1, the derivative of f(x) is just f'(x) and the derivative of 1 is 0

Answer 23

that is why the right hand side is
\[1+y'\]

Answer 24

so far so good?

Answer 25

yeah

Answer 26

ok now to the left hand side
\[(x-y)^2\]

Answer 27

which i will write as
\[(x-f(x))^2\]

Answer 28

by the chain rule the derivative of something squared is twice something times the derivative of something. so i write
\[2(x-f(x)) \times \frac{d}{dx}(x-f(x))\]

Answer 29

right

Answer 30

the the derivative of x is 1, the derivative of f(x) is f'(x) so we get
\[2(x-f(x))(1-f'(x))\] or
\[2(x-y)(1-y')\]

Answer 31

so we are back to
\[(2(x-y)(1-y')=1+y'\]and now our only job is to solve this equation for
\[y'\]

Answer 32

ugly algebra but we can work it out

Answer 33

\[2(x-xy'-y+yy')=1+y'\]
\[2x-2xy'-2y+2yy'=1+y'\]
\[2x-2y-1=y'+2xy'-2yy'\]

Answer 34

\[2x-2y-1=y'(1+2x-2y)\]
\[\frac{2x-2y-1}{1+2x-2y}=y'\]

Answer 35

ah shucks. i completely failed to realize that you could foil.

Answer 36

there might have been a snappier way to solve but i did not see it. also i could have made an algebra mistake so check my work

Answer 37

curses foiled again

Answer 38

haha.. thanks for taking the time to explain it!

Answer 39

really you should check my algebra because i just typed it out and didn't do it with pencil and paper

Answer 40

i am going to in a second. just going to do it on my notebook :P

Answer 41

hope it helped. good luck. are you done or do you have another one?

Answer 42

atm no. though i might skip to related rates and then yes.

Answer 43

related rates! must have just passed the chair rule part of the course. sliding ladders and oil spills, planes flying overhead and the guy walking away from the lamp. have fun!