Question

Implicit diff...where to go from here?
xcosy=1
So far I have:

Answer 1

notice that the left is a product, apply product rule

Answer 2

treat y as a function of x

Answer 3

x cosy=1

x' cosy + x cos'y = 1'

x' cosy - x siny y' = 1'

Answer 4

x' wrtx = 1

1' wrtx = 0

Answer 5

does that make sense?

Answer 6

Hang on

Answer 7

So where do I go from:
\[cosy-\sin(dy/dx)x=0\] .. after applying the Product Rule

Answer 8

thats not a correct application; your sin part is a little confuddled

Answer 9

cosy is a chain rule: -siny and out pops the derivative if the inner ... y'

Answer 10

cosy - x sin(y) y' = 0 then solve for y' by algebra

Answer 11

I think I have it, thanks

Answer 12

youre welcome :)