Expert: Sherry Wallin - 11/19/2008

Question
i have tried and i'm stuck, i dont know what kind of formula 2 use anymore.. I dont know when to apply product,quotient and chain rule as this topic required you to apply all, i dont know where to start. Can you help me? I'm doing my revision nw..Much appreciated..A) If x^2-y^2=1, prove that y(d^y/(dx)^2)+((dy)/(dx))^2=1, b)find the values of (dy)/(dx) & (d^2(y))/(dx^2) at the point (1,3) on the curve 3x^2+y^2=4y, c)if y^2-2xy=2x, prove that (x-y)((d^2y)/(dx^2)) + (2)((dy)/(dx))=((dy)/(dx))^2

Answer
I can help you with part b since I believe that problem is correctly written.
You are given 3x^2 y^2=4y and the first part wants you to evaluate y'=(dy/dx)
differentiate implicitly: 2*3x 2*y*y' = 4y'. You need to isolate y'
so subtract 2yy' and you will get 6x = 4y'-2yy', now factor the right hand side and get y'(4-2y) and divided both sides by (4-2y) and get 6x/(4-2y)= y', the left hand side can be reduced by a factor of 2 so you have 3x/(2-y) = y'. Now use your point and evaluate it using y' getting 3/-1, so y' is -3. To get y" just differentiate y' getting
y"=[(2-y)3 -3x(-y')]/(2-y)^2 = [(2-y)3 3xy']/(2-y)^2 You will use your point (1,3) and your new found value of y' = -3 and get
[(2-3)3  3(1)(-3)]/(2-3)^2 which equals
-1(3)-9/(-1)^2 = -12/1or just -12

I need some time to work part c, more later

Math Prof