Solving 2y' = csc2 (x-y): Is it Correct So Far?

[domyy]

Homework Statement

2y = cot(x-y)

The Attempt at a Solution

2y' = -csc² (x-y) . (1-y)(y')

Is it correct so far?

My book actually has

2y' = -csc^2 (x-y) + y' csc^2(x-y)

And I don't understand where that sum comes from. Am I supposed to apply the product rule? Because I tried and it didn't get me to that step the book presents.

 

[rock.freak667]

Your mistake would be in differentiating (x-y) wrt to x.

d/dx(x-y) = 1 - d/dx(y)

 

[domyy]

Thanks for the reply :)

But where does the sum sign come from?

Because even if it's 1 - y'

I will be still multiplying, as below:

-csc2 (x-y) . (1- y')

 

[Dick]

Thanks for the reply :)

But where does the sum sign come from?

Because even if it's 1 - y'

I will be still multiplying, as below:

-csc2 (x-y) . (1- y')

Multiply it out. Use that a(b+c)=ab+ac.