Question

@Loser66

Answer 1

@perl PLEASE HELPPPPPPPP

Answer 2

@freckles

Answer 3

what id derivative of (x+y) w.r.t x?

Answer 4

\[y=\sin(x+y) \\ y'=(x+y)'\cos(x+y) \text{ by chain rule } \]
So you need to find (x+y)'

Answer 5

1

Answer 6

(x+y)'=(x)'+(y)'
x'=1
but y' doesn't equal 0

Answer 7

y' is just y'

Answer 8

so (x+y)'=(x)'+(y)'=1+y'

Answer 9

No you need need to distribute
and solve for y'

Answer 10

d/dx(x) + d/dx(y) = 1 + dy/dx

Answer 11

now* not no

Answer 12

0

Answer 13

:(
where do you keep getting 0?

Answer 14

i didnt get 0 before??

Answer 15

then how did you get (x+y)' is 1?

Answer 16

i got 1

Answer 17

the derivative of (x+y) w.r.t to (x+y) is 1
but we are finding the derivative of (x+y) w.r.t x

Answer 18

im confused

Answer 19

\[\frac{d}{dx}(x+y) = \frac{d}{dx}(x)+\frac{d}{dx}y \text{ by sum rule } \\ \]

Answer 20

dx/dx is 1
dy/dx is y'

Answer 21

y'= cos(x+y)/ 1-cos(x+y)

Answer 22

you mean y'=cos(x+y)/[1-cos(x+y)] ?

Answer 23

and yes that would be correct

Answer 24

thanks

Answer 25

there are others way you can write it

Answer 26

if you want to anyways

Answer 27

if you are trying to show your answer is equivalent to some other answer listed