Solution to Midterm Problem 9.2

Example i: The problem was to use implicit differentiation to find y' in terms of x and y for
Step 1: Implicit differentiation. To the right you can see the steps, proceeding one by one, for each term, on how to apply implicit differentiation to this.

Step 2: An algebraic shuffle. Simply move all the terms that have a y' factor in them over to one side of the equation.

   3y²y' - 3xy'  =  3y - 4x

Step 3: Factor out the y'.

   y'(3y² - 3x)  =  3y - 4x

Step 4: Divide the other factor into sides, and you're done.

           3y - 4x
   y'  =           
          3(y² - x)

Example ii: Clearly here you have to use the quotient rule on this one. Your equation to differentiate is

x = - x² + y²

1 5

Step 1: Take the derivatives of the numerator and denominator. On the left you have a quotient with x² + y² in the denominator and x in the numerator. The derivative of the numerator is easy -- it's just 1. Now apply implicit differentiation to the denominator to see what its derivative is:

   d
     (x² + y²)  =  2x + 2yy'
   dx

Step 2: Apply the quotient rule. By exam time you should be able to recite this rule in your sleep. Our original equation has a quotient on the left and a constant on the right. So taking the derivative of the whole equation you get:

   (x² + y²)(1) - x(2x + 2yy')
                                =  0
            (x² + y²)²

Step 3: Look at the right-hand side of this result. To the right of the equal you've got zero. So if you multiply both sides by the denominator on the left, the (x² + y²)² goes away completely. Remember that when you have a quotient, only the numerator can make it be zero.

   (x² + y²)(1) - x(2x + 2yy')  =  0

Step 4: Algebraic shuffle again. You only have one term with a y' factor this time. So you get

   x² + y² - 2x²  =  2xyy'

Step 5: Divide out the remaining factor. Dividing both sides by 2xy and gathering like terms in the numerator, you get

    y² - x²
            =  y'
     2xy

Example iii: The problem was to use implicit differentiation to find y' in terms of x and y for

Step 1: Implicit differentiation. To the right you can see the steps, proceeding one by one, for each term, on how to apply implicit differentiation to this.

Step 2: Algebraic Shuffle. Again, move all the terms that have a y' factor to one side of the equation.

   y² - 6xy  =  3x²y' - 2xyy'

Step 3: Factor. Separate the y' from the other stuff on the right.

   y² - 6xy  =  y'(3x² - 2xy)

Step 4: Divide, and you're done.

    y² - 6xy
              =  y'
   3x² - 2xy

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