Expert: Sherman D. - 4/15/2009

Question
Find y' in terms of x and y by implicit differentiation if

1) xy+√(x+y)= 1

2) sin(x+2y)= cos(2x-y)  

Answer
(d/dx)(xy) + (d/dx)(sqrt(x + y)) = (d/dx)1

(xy)' = y + x(dy/dx) = y + 1

(d/dx)(sqrt(x + y)) =
(x + y)'(sqrt(x + y)') =
(1 + y')((1/2)(x + y)^(-1/2)) =
(1 + y')(1/(2sqrt(x + y)))
(1 + y')/(2sqrt(x + y))

so this gives you

(y + 1) + ((1 + y')/(2sqrt(x + y))) = 0

(1 + y')/(2sqrt(x + y)) = -(y + 1)
1 + y' = -2(y + 1)sqrt(x + y)
y' = -2(y + 1)sqrt(x + y) - 1

ANS : y' = (-2y - 2)sqrt(x + y) - 1

you can probably rewrite that other ways and still mean the same, but i'm leaving it at that.

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sin(x + 2y) = cos(2x - y)

sin(x + 2y)' = cos(2x - y)'

cos(x + 2y)(x + 2y)' = -sin(2x - y)(2x - y)'

cos(x + 2y)(1 + 2y') = -sin(2x - y)(2 - y')
cos(x + 2y) + 2y'cos(x + 2y) = -2sin(2x - y) + sin(2x - y)y'
-sin(2x - y)y' + 2cos(x + 2y)y' = -cos(x + 2y) - 2sin(2x - y)
y'(2cos(x + 2y) - sin(2x - y)) = -cos(x + 2y) - 2sin(2x - y)

y' = (-cos(x + 2y) - 2sin(2x - y))/(2cos(x + 2y) - sin(2x - y))

Not 100% on this one, so i recommend checking with answers.yahoo.com to see if they come up with the same thing before you rate me.