y=cosxcosy-sinxsiny/sinxcosy+cosxsiny
find dy/dx
plz show me working
I assume you meant:
y=(cosxcosy-sinxsiny)/(sinxcosy+cosxsiny)
= cos(x+y) / sin(x+y) , using your identities
= cot(x+y)
dy/dx = - csc^2 (x+y) (1 + dy/dx)
dy/dx = -csc^2 (x+y) - dy/dx csc^2 (x+y)
dy/dx + dy/dx csc^2 (x+y) = -csc^2 (x+y)
dy/dx(1 + csc^2 (x+y) ) = -csc^2 (x+y)
dy/dx = -csc^2 (x+y)/(1 + csc^2 (x+y) )
and, since y=cot(x+y),
dy/dx = -(y^2+1)/(y^2+2)
To find dy/dx, we need to differentiate y with respect to x, using the quotient rule:
dy/dx = (d/dx)(cos(x)cos(y) - sin(x)sin(y)) / (sin(x)cos(y) + cos(x)sin(y))
Let's differentiate the numerator first:
(d/dx)(cos(x)cos(y) - sin(x)sin(y)) = -sin(x)cos(y) - cos(x)sin(y)
Now, let's differentiate the denominator:
(d/dx)(sin(x)cos(y) + cos(x)sin(y)) = cos(x)cos(y) - sin(x)sin(y)
Now we can substitute these derivatives back into the original expression for dy/dx:
dy/dx = (-sin(x)cos(y) - cos(x)sin(y)) / (cos(x)cos(y) - sin(x)sin(y))
However, we can simplify this expression further. Let's multiply the numerator and denominator by -1:
dy/dx = (sin(x)cos(y) + cos(x)sin(y)) / (-cos(x)cos(y) + sin(x)sin(y))
The numerator and denominator both have the same terms but with opposite signs, so we can cancel them out:
dy/dx = 1
Therefore, dy/dx = 1.