1. MATH 31 REVIEWS
Chapter 2: Derivatives

2. Chapter 2: Derivatives
“ If you don't go off on a tangent while studying,
you can derive a great deal of success from it. ”

3. 1. Finding Derivatives from First Principles
Ex. If f(x) = 3x x , find f (x) using limits.

4. f(x) = 3x x + 1
f (x) = lim f(x+h) - f(x)
h0 h

5. f(x) = 3x x + 1
f (x) = lim f(x+h) - f(x)
h0 h
= lim [3(x+h) (x+h) + 1] - [3x2 - 4x + 1]
h0 h

6. f(x) = 3x x + 1
f (x) = lim f(x+h) - f(x)
h0 h
= lim [3(x+h) (x+h) + 1] - [3x2 - 4x + 1]
h0 h
= lim 3x2 + 6xh + 3h2 - 4x - 4h x2 + 4x - 1

7. f (x) = lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1

8. f (x) = lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1
= lim xh + 3h2 - 4h
h0 h

9. f (x) = lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1
= lim xh + 3h2 - 4h
h0 h
= lim h (6x + 3h - 4)

10. f (x) = lim 3x2 + 6xh + 3h2 - 4x - 4h + 1 - 3x2 + 4x - 1
= lim xh + 3h2 - 4h
h0 h
= lim h (6x + 3h - 4)
= lim (6x + 3h - 4) = 6x - 4
h0

11. 2. Sketching Derivative Functions
Remember, the derivative represents the tangent slope of a function.
Thus, the derivative function simply describes the slopes
of the original function.

12. e.g.
Sketch the derivative function y = f (x)
y = f(x)

13. y = f(x)
Horizontal slopes ...

14. y = f(x)
Horizontal slopes ...
y = f (x)
... become x-intercepts (f  = 0)

15. y = f(x)
Slope > 0
y = f (x)

16. y = f(x)
Slope > 0
y = f (x)
Positive y-coordinates

17. y = f(x)
Slope < 0
y = f (x)
Negative y-coordinates

18. Degree = 3
y = f(x)
y = f (x)
Degree = 2
For polynomial functions, the degree decreases by 1

19. 3. Power Rule
Differentiate y = 5x x
x x2 x3

20. y = 5x x
x x2 x3
y = 5x x x x x-3
Bring powers to the "top"

21. y = 5x x
x x2 x3
y = 5x x x x x-3
y = 5 (3x2) (-1x-2) x (-3x-4)
Ignore the coefficients
Differentiate the powers

22. y = 5x x
x x2 x3
y = 5x x x x x-3
y = 5 (3x2) (-1x-2) x (-3x-4)
y = 15x x x x-4

23. y = 5x x
x x2 x3
y = 5x x x x x-3
y = 5 (3x2) (-1x-2) x (-3x-4)
y = 15x x x x-4
y = 15x
x x x4

24. 4. Product Rule
Differentiate y = (4x x) (7x x ) using
the product rule. No need to simplify the answer.

25. y = (4x x) (7x x )
y = (4x3 + 9x) (7x4 - 11x2 - 3) + (4x3 + 9x) (7x4 - 11x2 - 3)
Product Rule: f  g + f g 

26. y = (4x x) (7x x )
y = (4x3 + 9x) (7x4 - 11x2 - 3) + (4x3 + 9x) (7x4 - 11x2 - 3)
y = (12x2 + 9) (7x4 - 11x2 - 3) + (4x3 + 9x) (28x3 - 22x)

27. 5. Quotient Rule
Differentiate y = x using the quotient rule.
x2 + 4x

28. y = x
x2 + 4x
y = (5 - 2x) (x2 + 4x) - (5 - 2x) (x2 + 4x)
(x2 + 4x)2
Quotient Rule: f  g  f g  g2

29. y = x
x2 + 4x
y = (5 - 2x) (x2 + 4x) - (5 - 2x) (x2 + 4x)
(x2 + 4x)2
y = (-2) (x2 + 4x) - (5 - 2x) (2x + 4)

30. y = x
x2 + 4x
y = (5 - 2x) (x2 + 4x) - (5 - 2x) (x2 + 4x)
(x2 + 4x)2
y = (-2) (x2 + 4x) - (5 - 2x) (2x + 4)
y = -2x2 - 8x - (10x x2 - 8x)

31. y = -2x2 - 8x - (10x x2 - 8x)
(x2 + 4x)2
y = -2x2 - 8x x x2 + 8x

32. y = -2x2 - 8x - (10x x2 - 8x)
(x2 + 4x)2
y = -2x2 - 8x x x2 + 8x
y = 2x x - 20
or y = 2(x2 - 5x - 10)
x2 (x + 4)2

33. 6. Chain Rule
Differentiate y = x - 9x2

34. y = x - 9x2
y = (4 + 3x - 9x2) 1/2

35. y = x - 9x2
y = (4 + 3x - 9x2) 1/2
y = (4 + 3x - 9x2) -1/2  d (4 + 3x - 9x2) dx
Derivative of the "outside function"
Don't forget to find the derivative of the "inside function"

36. y = x - 9x2
y = (4 + 3x - 9x2) 1/2
y = (4 + 3x - 9x2) -1/2  d (4 + 3x - 9x2) dx
y = (4 + 3x - 9x2) -1/2  (3 - 18x) 2

37. y = (4 + 3x - 9x2) -1/2  (3 - 18x) 2
y = x
x - 9x2 or
y = (1 - 6x)

38. 7. Combination of Rules
Where does the function y = (2x - 7)4 (3x + 1)6
have a horizontal tangent?

39. y = (2x - 7)4 (3x + 1)6
y = [(2x - 7)4] (3x + 1) (2x - 7)4 [(3x + 1)6]
Use the product rule first

40. y = (2x - 7)4 (3x + 1)6
y = [(2x - 7)4] (3x + 1) (2x - 7)4 [(3x + 1)6]
y = 4(2x - 7)3  (2)  (3x + 1)6 + (2x - 7)4  6(3x + 1)5  (3)
Use the chain rule next
Don't forget to do the derivative of the "inside function"

41. y = (2x - 7)4 (3x + 1)6
y = [(2x - 7)4] (3x + 1) (2x - 7)4 [(3x + 1)6]
y = 4(2x - 7)3  (2)  (3x + 1)6 + (2x - 7)4  6(3x + 1)5  (3)
= 8 (2x - 7)3 (3x + 1) (2x - 7)4 (3x + 1)5
Put both terms into the same "order"

42. y = (2x - 7)4 (3x + 1)6
y = [(2x - 7)4] (3x + 1) (2x - 7)4 [(3x + 1)6]
y = 4(2x - 7)3  (2)  (3x + 1)6 + (2x - 7)4  6(3x + 1)5  (3)
= 8 (2x - 7)3 (3x + 1) (2x - 7)4 (3x + 1)5
= 2 (2x - 7)3 (3x + 1)5 [ 4(3x + 1) + 9(2x - 7) ]
Factor out the common factors

43. y = (2x - 7)4 (3x + 1)6
y = [(2x - 7)4] (3x + 1) (2x - 7)4 [(3x + 1)6]
y = 4(2x - 7)3  (2)  (3x + 1)6 + (2x - 7)4  6(3x + 1)5  (3)
= 8 (2x - 7)3 (3x + 1) (2x - 7)4 (3x + 1)5
= 2 (2x - 7)3 (3x + 1)5 [ 4(3x + 1) + 9(2x - 7) ]
= 2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ]

44. y = 2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ]
When does the function have a horizontal tangent?

45. y = 2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ]
The function has a horizontal tangent when y = 0
2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ] = 0

46. y = 2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ]
The function has a horizontal tangent when y = 0
2 (2x - 7)3 (3x + 1)5 [ 30x - 32 ] = 0
when
2x - 7 = 0 or 3x + 1 = 0 or 30x - 32 = 0
x = x = x = 16

47. 8. Implicit Differentiation
Find y for the implicit relation x2y - 5x3 = y

48. 7. Implicit Differentiation
Find y for the implicit relation x2y - 5x3 = y
Remember, y is a function of x. So, you must treat it like
a separate function f(x).

49. x2y - 5x3 = y
(x2) y + x2  y - 15x2 = 4y3  y + 0
Product rule Chain rule

50. x2y - 5x3 = y
(x2) y + x2  y - 15x2 = 4y3  y + 0
2x y + x2 y - 15x2 = 4y3 y

51. x2y - 5x3 = y
(x2) y + x2  y - 15x2 = 4y3  y + 0
2x y + x2 y - 15x2 = 4y3 y
x2 y - 4y3 y = 15x xy
Get y terms on one side of the equations.
All others go on the other side.

52. x2y - 5x3 = y
(x2) y + x2  y - 15x2 = 4y3  y + 0
2x y + x2 y - 15x2 = 4y3 y
x2 y - 4y3 y = 15x xy
y (x2 - 4y3) = 15x2 - 2xy
Factor out the y

53. x2y - 5x3 = y
(x2) y + x2  y - 15x2 = 4y3  y + 0
2x y + x2 y - 15x2 = 4y3 y
x2 y - 4y3 y = 15x xy
y (x2 - 4y3) = 15x2 - 2xy
y = 15x2 - 2xy
x2 - 4y3