Expert: Sherry Wallin - 11/6/2009

Question
heres the problem:           cos(x+y) + sin(x-y)
heres how i started:

1. -sin(x+y)*(1 + dy/dx) + cos(x-y)*(1 - dy/dx) = 0

2.  dy/dx(sin(x+y) + cos(x-y)) = cos(x-y) - sin(x+y)

3.  dy/dx = (cos(x-y) - sin(x+y))/((sin(x+y) + cos(x-y))

my question regards step 2. How do u factor out that dy/dx? I am confused b/c (1+dy/dx) is still a function of the -sin. Also, i dnt clearly see how the cos(x-y) - sin((x+y) could be moved to the right w/o changing it sign. thanks you =]

Answer
Zachary~
   Here is how I would do the problem: I prefer to call dy/dx y' until the end
[cos(x+y)]' = -sin(x+y)*(1+y')=(-1-y')sin(x+y)= -sin(x+y)-y'sin(x+y)
[sin(x-y)]' = cos(x-y)*(1-y')=(1-y')cos(x-y)= cos(x-y)-y'cos(x-y)
so [cos(x+y) + sin(x-y)]' = -sin(x+y)-y'sin(x+y)+cos(x-y)-y'cos(x-y)

= -y'[sin(x+y)+cos(x-y)]+ cos(x-y) - sin(x+y)=(-dy/dx)[sin(x+y)+cos(x-y)]+ cos(x-y) - sin(x+y)

Were you given cos(x+y) + sin(x-y) = 0 to begin with? You didn't write that as your problem. In fact you didn't state that you were suppose to solve for y' = dy/dx. Based on how you gave me the problem the best you can do is what I have done above. When not given an equation, you cannot solve for anything. In other words you need to be given two expressions equal to each other in order to 'solve' for anything. You must be given an 'equal' sign in order to 'solve' for anything!

Write me again if this doesn't help you figure out what you need to do.

Math Prof