Expert: Sherry Wallin - 11/21/2008

Question
i'm stuck with this question, i'm not really understood this topic, i dont know when to apply this formula when to apply that formula, i dont know why it become like that after differentiate as i see the textbook example as the textbook didnt explain it. I'm revising now. Please can you help me with this question? Much obliged. Find the values of y" at the point (1,-2) on the curve 3(x^2)+2xy+y^2=3

Answer
First you need to find the 2nd derivative y". By the way, it is called implicit differentiation because you have two variables and neither is solved for.

Ok let me explain to you what I do at each step
Step 1:  the derivative of 3x^2 is 6x
Step 2: to take the derivative of 2xy you need to use the product rule: take the first term (2x) times the derivative of the second term(y') plus the derivative of the first term 2 times the second term y
so for the derivative of 2xy you have  2xy'+ 2y
Step 3: take the derivative of y^2 which is 2y*y' (the times y' is the implicit part). And
Step 4: the derivative of a constant 3 is 0
so what you have is 6x +  2xy'+ 2y + 2yy' = 0
Step 5: solve for y' you will need this near the end
2xy'+2yy' = -6x - 2y, factor the y' out and get: y'(2x + 2y) = -6x -2y
divide by 2x + 2y and get y' = (-6x -2y)/(2x + 2y) and finally factor out a 2 and cancel leaving y' = -2(3x+y)/2(x+y) = -(3x+y)/(x+y)
In order to calculate y" you need to take the derivative again. Notice on the left you have y' and it's derivative is y". Now take the derivative on the right side doing the following:
Step 6:use the quotient rule: bottom times derivative of the top minus the top times the derivative of the bottom all over the bottom squared: [(x+y)(-3-y')+(3x+y)(1+y')]/(x+y)^2 = 2(xy'-y)/(x+y)^2. You need to repace y' with -(3x+y)/(x+y) and you get:
y" =  -2(3x^2+2xy+y^2)/(x+y)^3
now plug in your point (1, -2) and get 6
y" = -2(3*1+2*1*(-2)+(-2)^2)/(1-2)^3
  = -2(3-4+4)/-1
  = -2(3)/-1
  = 6
Hopefully this has been explained to you in a way in which you can understand and learn from.

Math Prof