1. I chose to use implicit differentiation. Product Rule u’v + uv’
Part (a)
I chose to use implicit differentiation.
Product Rule u’v + uv’
x2 + 4y2 = x y
2x + 8y(dy/dx) = 3y + 3x(dy/dx)
8y(dy/dx) - 3x(dy/dx) = 3y – 2x
Factor out a (dy/dx)
(dy/dx) (8y - 3x) = 3y – 2x
(dy/dx) =
3y – 2x
8y – 3x

2. They’re telling us to set the derivative equal to zero.
Part (b)
They’re telling us to set the derivative equal to zero.
(dy/dx) =
3y – 2x
8y – 3x
From Part (a)
3y – 2x
8y – 3x
= 0
Cross multiply
3y – 2x = 0
y = 2
3y – 6 = 0
There’s a horizontal tangent at (3,2)

3. There’s a horizontal tangent at (3,2)
Part (b)
x2 + 4y2 = xy
32 + (4)(2)2 = (3)(2) =
25 = Yup!
It would be a good idea to make sure that (3,2) is actually on our curve!!!
There’s a horizontal tangent at (3,2)

4. Quotient Rule (u’v - uv’)/v2 u’ = 3(dy/dx) – 2 v’ = 8(dy/dx) - 3
Part (c)
First, we’re being asked to find the value of the 2nd derivative at (3,2).
(dy/dx) =
3y – 2x
8y – 3x
Quotient Rule (u’v - uv’)/v2
u’ = 3(dy/dx) – 2 v’ = 8(dy/dx) - 3
d2y/dx2 =
[3(dy/dx) – 2](8y – 3x) - (3y – 2x) [8(dy/dx) – 3]
(8y – 3x)2

5. First, we’re asked to find the value of the 2nd derivative at (3,2).
Part (c)
First, we’re asked to find the value of the 2nd derivative at (3,2).
Keep in mind that the value of the 1st derivative at (3,2) is zero.
d2y/dx2 =
[3(dy/dx) – 2](8y – 3x) - (3y – 2x) [8(dy/dx) – 3]
(8y – 3x)2
d2y/dx2 =
[3(0) – 2](8y – 3x) - (3y – 2x) [8(0) – 3]
(8y – 3x)2

6. Part (c) d2y/dx2 = = d2y/dx2 = @ (3,2) d2y/dx2 = (8y – 3x)2
(–2)(7) – (0)(-3)
(7)2 7 -2
d2y/dx2 =
(–2)[(8)(2) – 3(3)] – [(3)(2) – (2)(3)](–3)
[(8)(2) – (3)(3)]2
@ (3,2)
d2y/dx2 =
[3(0) – 2](8y – 3x) - (3y – 2x) [8(0) – 3]
(8y – 3x)2

7. Part (c)
d2y/dx2 = =
(–2)(7) – (0)(-3)
(7)2 7 -2
The 1st derivative is zero at (3,2) while the 2nd derivative is negative there. Therefore, according to the 2nd derivative test, the curve has a local maximum at (3,2).