Expert: Ahmed Salami - 12/11/2008

Question
QUESTION: I need help implicitly differentiating cos(x+y) + sin (x-y) = 0 @ y' = (pi/4, -pi/4). There is something going on here that I'm not seeing. Is there a substitution that I need to make?

ANSWER: Hi Marcel,
I think you mean y' at (#/4, -#/4)
where # represents pi
What you need to do is first determine the expression for y' (i.e dy/dx).
If we let z = cos(x+y) + sin(x-y), then
dy/dx = -(&z/&x)/(&z/&y)    (partial differentials)
but
&z/&x = -sin(x+y) + cos(x-y)
     = cos(x-y) - sin(x+y)
&z/&y = -sin(x+y) + -cos(x-y)
     = -[cos(x-y) + sin(x+y)]
Therefore,
y' = dy/dx
  = -[cos(x-y) - sin(x+y)]/-[cos(x-y) + sin(x+y)]
  = [cos(x-y) - sin(x+y)]/[cos(x-y) + sin(x+y)]
At (#/4, -#/4)
y' = [cos(#/4 - -#/4) - sin(#/4 + -#/4)]/[cos(#/4 - -#/4) + sin(#/4 + -#/4)]
  = [cos(#/4 + #/4) - sin(#/4 - #/4)]/[cos(#/4 + #/4) + sin(#/4 - #/4)]
  = [cos(#/2) - sin(0)]/[cos(#/2) + sin(0)]
  = [cos(#/2) - sin(0)]/[cos(#/2) + sin(0)]
  = 0 - 0 / 0 + 0
  = 0/0 (undefined)

Regards

---------- FOLLOW-UP ----------

QUESTION: Can we try again? I'm being given the equation for the tangent line as Tangent is y = - 3(x - 1). (4) (1 + y')cos(x + y) - (1 - y')sin(x - y) = 0 and y= -pi/4. See this website section for clarity http://www.maths.abdn.ac.uk/~igc/tch/ma1002/diff/node81.html

Answer
Hi Marcel,
You've obviously been mixing things up. First you send me a different question from the one on that page by interchanging the sine and cosine
variables.
The result on the page states
(1 + y')cos(x+y) - (1 - y')sin(x-y) = 0
and doesnt include y = - 3(x - 1). (4) which you've again put mistakenly!!!
Now, i would explain again (the right question).
If we let z = sin(x+y) + cos(x-y), then
dy/dx = -(&z/&x)/(&z/&y)    (partial differentials)
but
&z/&x = cos(x+y) - sin(x-y)
&z/&y = cos(x+y) + -sin(x-y)(-1)
     = cos(x+y) + sin(x-y)
Therefore,
y' = dy/dx
 = -[cos(x+y) - sin(x-y)]/[cos(x+y) + sin(x-y)]
 = [sin(x-y) - cos(x+y)]/[sin(x-y) + cos(x+y)]
You get the same result when you make y' subject of the formula in
(1 + y')cos(x+y) - (1 - y')sin(x-y) = 0

At (#/4, -#/4)
y' = [sin(#/4 - -#/4) - cos(#/4 + -#/4)]/[sin(#/4 - -#/4) + cos(#/4 + -#/4)]
 = [sin(#/4 + #/4) - cos(#/4 - #/4)]/[sin(#/4 + #/4) + cos(#/4 - #/4)]
 = [sin(#/2) - cos(0)]/[sin(#/2) + cos(0)]
 = 1 - 1 / 1 + 1
 = 0/2
 = 0
y' = 0 means that the tangent line at that point is horizontal and is
y = -#/4

Note that the derivative as gotten on the page goes thus,
sin(x+y) + cos(x-y) = 0
d/dx[sin(x+y)] + d/dx[cos(x-y)] = d/dx[0]
cos(x+y)d/dx(x+y) + -sin(x-y)d/dx(x-y) = 0
cos(x+y)(1 + y') + -sin(x-y)(1 - y') = 0
(1 + y')cos(x+y) - (1 - y')sin(x-y) = 0
Faster, but needs more understanding.

Regards