### File

```( x  2 )( x
2
3x 
2
2

x
1
5
x
2
3x  2 x  4
4
3
2x
2
 1)
Day #1 Homework
page 124: 1-13, 15-22 (do not show
graphically)
If is the function with the constant value c,
then df/dx = d/dx (c) = 0
Example: y = 4, find dy/dx
dy/dx= 0
If n is a positive integer, then d/dx (xn) = nxn-1
Examples:
 y = x3
 dy/dx= 3x2
 y = 4x4
 dy/dx = 16x3
If u and v are differentiable functions of x,
then their sum and differences are
differentiable at every point where u and v
are differentiable:
d/dx ( u ± v) = du/dx ± dv/dx
Example:
 y = x3 + ½ x2 – 3x + 2
 dy/dx= 3x2 + x - 3
Does the curve y = x4 – 2x2 + 2 have any
horizontal tangents? If so, where?
 Calculate dy/dx
 4x3 – 4x
 Solve the equation dy/dx = 0 for x.
 4x3 – 4x = 0
 4x (x2 – 1) = 0
 4x (x – 1) (x + 1) = 0
 4x = 0; (x – 1) = 0; (x + 1) = 0
 x = 0, 1, -1
d/dx (uv) = u (dv/dx) + v(du/dx)
Find f’(x) if f(x) = (x2 + 1) (x3 + 3)
Let u = x2 + 1 and v = x3 + 3
dv/dx = 3x2 and du/dx = 2x
 u (dv/dx) + v(du/dx) = (x2 + 1 )(3x2) + (x3 + 3)(2x)
3x4 + 3x2 + 2x4 + 6x
5x4 + 3x2 + 6x
v
d u
 
dx  v 
du
u
dx
v
2
2
2x  2x  2x  2x
3
v  x 1
2
3
( x  1)
2
dv / dx  2 x
v  ( x  1)
x 1
2
d  u  ( x  1 ) 2 x  ( x  1) 2 x
 
2
2
dx  v 
( x  1)
du / dx  2 x
2
f ( x) 
2
2
x 1
2
dx
u  x 1
2
dv
 4x
2
( x  1)
2
2
2
6
2x
3
(x
–
4
5x
+ x)
+ 2x
2
(2x
+ 3)
x 1
3
x
2
1
Let y1 = f’(x) and y2 = NDER f(x). They should
coincide.
Let y = uv be the product of the functions u
and v. Find y’(2) if
 u(2) = 3, u’(2) = -4, v(2) = 1 and v’(2) = 2
 So, from the product rule:
 u(2)v’(2) + v(2)u’(2)
 3(2) + 1(-4) = 2
If n is a negative integer and x ≠ 0, then
 d/dx (xn) = nxn-1
 Example: f(x) = -2x-3
 f’(x) = 6x-4
Find an equation for the line tangent to the
curve y = (x2 + 3)/2x at the point (1, 2)
 While we could find the derivative by the Quotient
Rule, but it is easier to first simplify into a sum of two
powers: (x2/2x) + 3/2x = ( ½ ) x + (3/2)x-1
 dy/dx = ½ - (3/2)x-2
 The slope at x = 1:
 ½ - (3/2) (1)-2 =
½ - 3/2 = -1
 Using the slope of -1 and the point (1, 2):
 y – 2 = -1 (x – 1 )  y = -1x + 3
dy/dx: first derivative
 f’(x) = y’
dy’/dx = second derivative
 f’’(x) = y’’ = d2y/dx2
dy’’/dx = third derivative
 f’’’(x) = y’’’ = d3y/dx3
etc….
y’: 3x2 – 10x
y’’ = 6x – 10
y’’’ = 6
y’’’’= 0
(page 123, example 5)
An orange farmer currently has 200 trees
yielding an average of 15 bushels of oranges
per tree. She is expanding her farm at the rate
of 15 trees per year, while improved
husbandry is improving her average annual
yield by 1.2 bushels by tree. What is the
current (instantaneous) rate of increase of her
total annual production of oranges?
 Let t(x) = the number of trees x years from now
 Let y(x) = yield per tree x years from now
 p(x) = t(x)y(x) is the total production of oranges in
year x.
 We know t(0) = 200; y (0) = 15; t’(0) = 15; y’(0) = 1.2
 We need to find p’(x), so we can use the product rule:
 t(0)y’(0) + y(0)t’(0)
 200(1.2) + 15(15) = 465 bushels per year.
page 124-125:23-37, 51
```