Given: #x sqrt(1+y) + y sqrt(1+x) = 0#
We need to use implicit differentiation. We will use these differentiation rules for each segment of the equation:
The product rule: #(u*v)' = u*v' + v*u'#
The power rule: #(u^n)' = n u^(n-1) u'#
#(0)' = 0#
First segment of the equation:
Let #u = x; " "u' = 1#
#" "v = sqrt(1+y) = (1+y)^(1/2); " " v' = 1/2(1+y)^(-1/2)y'#
#" "v' = (y')/(2sqrt(1+y)#
Second segment of the equation:
Let #u = y; " "u' = y'#
#" "v = sqrt(1+x) = (1+x)^(1/2); " " v' = 1/2(1+x)^(-1/2)(1)#
#" "v' = 1/(2sqrt(1+x)#
Putting it all together using the product rule:
#(xy')/(2sqrt(1+y)) + sqrt(1+y) + y/(2sqrt(1+x)) +y' sqrt(1+x)= 0#
Keep all terms that contain #y'# on the left side of the equation:
#(xy')/(2sqrt(1+y)) +y' sqrt(1+x) = -sqrt(1+y) -y/(2sqrt(1+x))#
Factor #y'#;
#y' (x/(2sqrt(1+y)) + sqrt(1+x)) = -(sqrt(1+y) +y/(2sqrt(1+x)))#
Isolate #y'# using division:
#y' = -(sqrt(1+y) +y/(2sqrt(1+x)))/(x/(2sqrt(1+y)) + sqrt(1+x))#
This answer can be simplified if needed by finding common denominators in both the numerator and denominator.