Expert: Paul Klarreich - 12/7/2008

Question
cos(x+y) + sin (x-y) = 0 @ (pi/4, -pi/4)
The answer is
y = - 3(x - 1). (4) (1 + y')cos(x + y) - (1 - y')sin(x - y) = 0
I come close with cos(x+y) +y'cos (x+y)-sin(x-y) +y'sin(x-y).
Can you fill me in or give me hint on what to do with the rest of it?

Answer
Questioner:   Marcel Lane
Category:  Calculus
Private:  No
 
Subject:  implicit differentation
Question:  cos(x+y) + sin (x-y) = 0 @ (pi/4, -pi/4)
The answer is
y = - 3(x - 1). (4) (1 + y')cos(x + y) - (1 - y')sin(x - y) = 0
I come close with cos(x+y) +y'cos (x+y)-sin(x-y) +y'sin(x-y).
Can you fill me in or give me hint on what to do with the rest of it?
..........................
Hi, Marcel,

The first thing you have to do is ask a question.
I assume you meant to write:

If cos(x+y) + sin (x-y) = 0, find dy/dx @ (pi/4, -pi/4).

You are going to have trouble here.  There is this little thing -- the given point is supposed to be ON THE GRAPH -- it is supposed to solve the equation.  but:

cos(pi/4+ (-pi/4)) + sin (pi/4 -(-pi/4)) = 0

cos(0) + sin (pi/2) = 0

1 + 1 = 0

is not true.  

Take another look.