The document discusses implicit equations and implicit differentiation. It begins by explaining the difference between explicit and implicit forms of equations, using the example of y=1/x which can be written explicitly as y=1/x or implicitly as xy=1. It then introduces the concept of implicit differentiation, which involves taking the derivative of an implicit equation with respect to x and solving for the derivative of y with respect to x (y’). This allows one to find the slope of the curve at a point, even if the explicit form of the relation between x and y is difficult to determine from the implicit equation.

1. Implicit Equations and Implicit Differentiation

2. Implicit Equations and Implicit Differentiation
Implicit Equations

3. Implicit Equations and Implicit Differentiation
Implicit Equations
The function y = 1/x may be written as xy = 1.

4. Implicit Equations and Implicit Differentiation
Implicit Equations
The function y = 1/x may be written as xy = 1.
The y = 1/x is the explicit form because the output y is
given explicitly as a function of x. The equation
xy = 1 is said to be in the implicit form because the
relation between x and y are not given as a function.

5. Implicit Equations and Implicit Differentiation
Implicit Equations
The function y = 1/x may be written as xy = 1.
The y = 1/x is the explicit form because the output y is
given explicitly as a function of x. The equation
xy = 1 is said to be in the implicit form because the
relation between x and y are not given as a function.
The implicit equation x2 + y2 = 1 has the unit circle as
its graph.

6. Implicit Equations and Implicit Differentiation
Implicit Equations
The function y = 1/x may be written as xy = 1.
The y = 1/x is the explicit form because the output y is
given explicitly as a function of x. The equation
xy = 1 is said to be in the implicit form because the
relation between x and y are not given as a function.
The implicit equation x2 + y2 = 1 has the unit circle as
its graph. It’s not one function. y = √1 – x2
However x2 + y2 = 1
may be represented by
two different functions.
y = √1 – x2 or y = –√1 – x2
y = – √1 – x2

7. Implicit Equations and Implicit Differentiation
The joint variables x and y in an implicit equation may
be inseparable by algebra.

8. Implicit Equations and Implicit Differentiation
The joint variables x and y in an implicit equation may
be inseparable by algebra. sin(xy) = 0,

9. Implicit Equations and Implicit Differentiation
The joint variables x and y in an implicit equation may
be inseparable by algebra. In fact for sin(xy) = 0,
there are infinitely many solution-branches xy = nπ
for n = 0, 1, 2,..

10. Implicit Equations and Implicit Differentiation
The joint variables x and y in an implicit equation may
be inseparable by algebra. In fact for sin(xy) = 0,
there are infinitely many solution-branches xy = nπ
for n = 0, 1, 2,.. Plot xy = 0, xy = π, xy = – π, ..

11. Implicit Equations and Implicit Differentiation
The joint variables x and y in an implicit equation may
be inseparable by algebra. In fact for sin(xy) = 0,
there are infinitely many solution-branches xy = nπ
for n = 0, 1, 2,.. Plot xy = 0, xy = π, xy = – π, ..
y
x
sin(xy) = 0

12. Implicit Equations and Implicit Differentiation
The joint variables x and y in an implicit equation may
be inseparable by algebra. In fact for sin(xy) = 0,
there are infinitely many solution-branches xy = nπ
for n = 0, 1, 2,.. Plot xy = 0, xy = π, xy = – π, ..
Any point (x, y) on any of these branch is a solution
for sin(xy) = 0.
y
xy = 0 x
sin(xy) = 0

13. Implicit Equations and Implicit Differentiation
The joint variables x and y in an implicit equation may
be inseparable by algebra. In fact for sin(xy) = 0,
there are infinitely many solution-branches xy = nπ
for n = 0, 1, 2,.. Plot xy = 0, xy = π, xy = – π, ..
Any point (x, y) on any of these branch is a solution
for sin(xy) = 0.
y
xy = 0 x
xy = –π
sin(xy) = 0

14. Implicit Equations and Implicit Differentiation
The joint variables x and y in an implicit equation may
be inseparable by algebra. In fact for sin(xy) = 0,
there are infinitely many solution-branches xy = nπ
for n = 0, 1, 2,.. Plot xy = 0, xy = π, xy = – π, ..
Any point (x, y) on any of these branch is a solution
for sin(xy) = 0.
y
xy = 0 x
xy = –π xy = –π
sin(xy) = 0

15. Implicit Equations and Implicit Differentiation
The joint variables x and y in an implicit equation may
be inseparable by algebra. In fact for sin(xy) = 0,
there are infinitely many solution-branches xy = nπ
for n = 0, 1, 2,.. Plot xy = 0, xy = π, xy = – π, ..
Any point (x, y) on any of these branch is a solution
for sin(xy) = 0.
Given a point P(a, b) that satisfies y
a given implicit equation, P must
be on one (or more ) of the
branch which is well defined by
some function y = y(x). xy = 0 x
xy = –π xy = –π
sin(xy) = 0

16. Implicit Equations and Implicit Differentiation
The joint variables x and y in an implicit equation may
be inseparable by algebra. In fact for sin(xy) = 0,
there are infinitely many solution-branches xy = nπ
for n = 0, 1, 2,.. Plot xy = 0, xy = π, xy = – π, ..
Any point (x, y) on any of these branch is a solution
for sin(xy) = 0.
Given a point P(a, b) that satisfies y
a given implicit equation, P must
be on one (or more ) of the
branch which is well defined by y=π/x
P(π, 1)
some function y = y(x). x
For example, the point (π, 1)
satisfies sin(xy) = 0 and it’s on the
branch xy = π or y = y(x) = π/x. sin(xy) = 0

17. Implicit Equations and Implicit Differentiation
We assume that the branch y
function y = y(x) that contains P
is differentiable at the point P as
P(π, 1)
the case in the example here y=π/x
that y = π/x is differentiable at x
P(π, 1).
sin(xy) = 0

18. Implicit Equations and Implicit Differentiation
We assume that the branch y
function y = y(x) that contains P
is differentiable at the point P as
P(π, 1)
the case in the example here y=π/x
that y = π/x is differentiable at x
P(π, 1). The slope at P may be
found by taking the derivative of
the branch function y = π/x sin(xy) = 0
which may be difficult to identify.

19. Implicit Equations and Implicit Differentiation
We assume that the branch y
function y = y(x) that contains P
is differentiable at the point P as
P(π, 1)
the case in the example here y=π/x
that y = π/x is differentiable at x
P(π, 1). The slope at P may be
found by taking the derivative of
the branch function y = π/x sin(xy) = 0
which may be difficult to identify.
Instead we may compute a different form of the
derivative directly from the implicit equation sin(xy) = 0
and use it to compute the slope at the point P(π, 1).

20. Implicit Equations and Implicit Differentiation
We assume that the branch y
function y = y(x) that contains P
is differentiable at the point P as
P(π, 1)
the case in the example here y=π/x
that y = π/x is differentiable at x
P(π, 1). The slope at P may be
found by taking the derivative of
the branch function y = π/x sin(xy) = 0
which may be difficult to identify.
Instead we may compute a different form of the
derivative directly from the implicit equation sin(xy) = 0
and use it to compute the slope at the point P(π, 1).
This procedure is called implicit differentiation and the
form of the derivative is called the implicit derivative.

21. Implicit Equations and Implicit Differentiation
To differentiate implicitly, we assume that y is y(x)
a function of x, and that we are to solve for y'(x).

22. Implicit Equations and Implicit Differentiation
To differentiate implicitly, we assume that y is y(x)
a function of x, and that we are to solve for y'(x).
Because y is y(x), all the the y’s in an implicit
equation are subjected to the Chain Rule.

23. Implicit Equations and Implicit Differentiation
To differentiate implicitly, we assume that y is y(x)
a function of x, and that we are to solve for y'(x).
Because y is y(x), all the the y’s in an implicit
equation are subjected to the Chain Rule.
Example A. Assume that y is y(x), expand the
following derivative in expressions of x, y and y'.
a. (x3 y)'
b. (x y3)'

24. Implicit Equations and Implicit Differentiation
To differentiate implicitly, we assume that y is y(x)
a function of x, and that we are to solve for y'(x).
Because y is y(x), all the the y’s in an implicit
equation are subjected to the Chain Rule.
Example A. Assume that y is y(x), expand the
following derivative in expressions of x, y and y'.
a. (x3 y)'
By the Product Rule,
(x3 y)' = (x3)' y + x3 (y)'
b. (x y3)'

25. Implicit Equations and Implicit Differentiation
To differentiate implicitly, we assume that y is y(x)
a function of x, and that we are to solve for y'(x).
Because y is y(x), all the the y’s in an implicit
equation are subjected to the Chain Rule.
Example A. Assume that y is y(x), expand the
following derivative in expressions of x, y and y'.
a. (x3 y)'
By the Product Rule,
(x3 y)' = (x3)' y + x3 (y)' = 3x2y + x3(y)'
b. (x y3)'

26. Implicit Equations and Implicit Differentiation
To differentiate implicitly, we assume that y is y(x)
a function of x, and that we are to solve for y'(x).
Because y is y(x), all the the y’s in an implicit
equation are subjected to the Chain Rule.
Example A. Assume that y is y(x), expand the
following derivative in expressions of x, y and y'.
a. (x3 y)'
By the Product Rule,
(x3 y)' = (x3)' y + x3 (y)' = 3x2y + x3(y)'
b. (x y3)'
By the Product Rule,
(x y3)' = (x)' y3 + x3 (y3)'

27. Implicit Equations and Implicit Differentiation
To differentiate implicitly, we assume that y is y(x)
a function of x, and that we are to solve for y'(x).
Because y is y(x), all the the y’s in an implicit
equation are subjected to the Chain Rule.
Example A. Assume that y is y(x), expand the
following derivative in expressions of x, y and y'.
a. (x3 y)'
By the Product Rule,
(x3 y)' = (x3)' y + x3 (y)' = 3x2y + x3(y)'
b. (x y3)'
(3y2y')
By the Product Rule,
(x y3)' = (x)' y3 + x3 (y3)'

28. Implicit Equations and Implicit Differentiation
To differentiate implicitly, we assume that y is y(x)
a function of x, and that we are to solve for y'(x).
Because y is y(x), all the the y’s in an implicit
equation are subjected to the Chain Rule.
Example A. Assume that y is y(x), expand the
following derivative in expressions of x, y and y'.
a. (x3 y)'
By the Product Rule,
(x3 y)' = (x3)' y + x3 (y)' = 3x2y + x3(y)'
b. (x y3)'
(3y2y')
By the Product Rule,
(x y3)' = (x)' y3 + x3 (y3)' = y3 + 3x3y2y'

29. Implicit Equations and Implicit Differentiation
d. Expand (x3 y – x y3)' and collect the y' – term

30. Implicit Equations and Implicit Differentiation
d. Expand (x3 y – x y3)' and collect the y' – term
(x3 y – x y3)'
= [3x2y + x3 y'] – [y3 + 3x3y2 y']

31. Implicit Equations and Implicit Differentiation
d. Expand (x3 y – x y3)' and collect the y' – term
(x3 y – x y3)'
= [3x2y + x3 y'] – [y3 + 3x3y2 y']
= 3x2y + x3 y' – y3 – 3x3y2 y'

32. Implicit Equations and Implicit Differentiation
d. Expand (x3 y – x y3)' and collect the y' – term
(x3 y – x y3)'
= [3x2y + x3 y'] – [y3 + 3x3y2 y']
= 3x2y + x3 y' – y3 – 3x3y2 y'
= 3x2y – y3 + (x3 – 3x3y2) y'

33. Implicit Equations and Implicit Differentiation
d. Expand (x3 y – x y3)' and collect the y' – term
(x3 y – x y3)'
= [3x2y + x3 y'] – [y3 + 3x3y2 y']
= 3x2y + x3 y' – y3 – 3x3y2 y'
= 3x2y – y3 + (x3 – 3x3y2) y'

34. Implicit Equations and Implicit Differentiation
d. Expand (x3 y – x y3)' and collect the y' – term
(x3 y – x y3)'
= [3x2y + x3 y'] – [y3 + 3x3y2 y']
= 3x2y + x3 y' – y3 – 3x3y2 y'
= 3x2y – y3 + (x3 – 3x3y2) y'
y
Let us go back to the
equation sin(xy) = 0.
x
sin(xy) = 0

35. Implicit Equations and Implicit Differentiation
d. Expand (x3 y – x y3)' and collect the y' – term
(x3 y – x y3)'
= [3x2y + x3 y'] – [y3 + 3x3y2 y']
= 3x2y + x3 y' – y3 – 3x3y2 y'
= 3x2y – y3 + (x3 – 3x3y2) y'
y
Let us go back to the
equation sin(xy) = 0.
P(π, 1)
Example B. Use implicit
y=π/x
differentiation to find y' x
given that sin(xy) = 0 and find
the slope and the tangent line
at the point (π, 1) on the graph sin(xy) = 0
of sin(xy) = 0.

36. Implicit Equations and Implicit Differentiation
Differentiate both sides of sin(xy) = 0 with respect to x,
[sin(xy) ] ' = (0)'
y
P(π, 1)
y=π/x
x
sin(xy) = 0

37. Implicit Equations and Implicit Differentiation
Differentiate both sides of sin(xy) = 0 with respect to x,
[sin(xy) ] ' = (0)' by the Sine Chain Rule
[cos(xy)] (xy)' = 0
y
P(π, 1)
y=π/x
x
sin(xy) = 0

38. Implicit Equations and Implicit Differentiation
Differentiate both sides of sin(xy) = 0 with respect to x,
[sin(xy) ] ' = (0)' by the Sine Chain Rule
[cos(xy)] (xy)' = 0 by the Product Rule
cos(xy) [(x)'y + x(y)'] = 0 y
P(π, 1)
y=π/x
x
sin(xy) = 0

39. Implicit Equations and Implicit Differentiation
Differentiate both sides of sin(xy) = 0 with respect to x,
[sin(xy) ] ' = (0)' by the Sine Chain Rule
[cos(xy)] (xy)' = 0 by the Product Rule
cos(xy) [(x)'y + x(y)'] = 0 y
cos(xy) [1*y + x(y)'] = 0
P(π, 1)
y=π/x
x
sin(xy) = 0

40. Implicit Equations and Implicit Differentiation
Differentiate both sides of sin(xy) = 0 with respect to x,
[sin(xy) ] ' = (0)' by the Sine Chain Rule
[cos(xy)] (xy)' = 0 by the Product Rule
cos(xy) [(x)'y + x(y)'] = 0 y
cos(xy) [1*y + x(y)'] = 0
P(π, 1)
Solve for (y)', set y + x(y)' = 0
y=π/x
x
sin(xy) = 0

41. Implicit Equations and Implicit Differentiation
Differentiate both sides of sin(xy) = 0 with respect to x,
[sin(xy) ] ' = (0)' by the Sine Chain Rule
[cos(xy)] (xy)' = 0 by the Product Rule
cos(xy) [(x)'y + x(y)'] = 0 y
cos(xy) [1*y + x(y)'] = 0
P(π, 1)
Solve for (y)', set y + x(y)' = 0
y=π/x
–y x
x(y)' = –y or that y' = x
sin(xy) = 0

42. Implicit Equations and Implicit Differentiation
Differentiate both sides of sin(xy) = 0 with respect to x,
[sin(xy) ] ' = (0)' by the Sine Chain Rule
[cos(xy)] (xy)' = 0 by the Product Rule
cos(xy) [(x)'y + x(y)'] = 0 y
cos(xy) [1*y + x(y)'] = 0
P(π, 1)
Solve for (y)', set y + x(y)' = 0
y=π/x
–y x
x(y)' = –y or that y' = x
At P(π, 1), x = π, y = 1
y' = –1 = the slope at P.
π sin(xy) = 0

43. Implicit Equations and Implicit Differentiation
Differentiate both sides of sin(xy) = 0 with respect to x,
[sin(xy) ] ' = (0)' by the Sine Chain Rule
[cos(xy)] (xy)' = 0 by the Product Rule
cos(xy) [(x)'y + x(y)'] = 0 y
cos(xy) [1*y + x(y)'] = 0
P(π, 1)
Solve for (y)', set y + x(y)' = 0
y=π/x
–y x
x(y)' = –y or that y' = x
At P(π, 1), x = π, y = 1
y' = –1 = the slope at P.
π sin(xy) = 0
Using the the Point–Slope Formula at P (π, 1), the
equation of the tangent line is y = –x + 2.
π

44. Implicit Equations and Implicit Differentiation
Your turn: Expand [cos3 (x/y)] '

45. Implicit Equations and Implicit Differentiation
Your turn: Expand [cos3 (x/y)] '
By the Power Chain Rule,
[cos3 (x/y)] ' = 3cos2(x/y) [cos (x/y)] '
= 3cos2(x/y) [– sin(x/y)] (x/y) '
= 3cos2(x/y) [– sin(x/y)] ((y – xy')/y2)

46. Implicit Equations and Implicit Differentiation
Your turn: Expand [cos3 (x/y)] '
By the Power Chain Rule,
[cos3 (x/y)] ' = 3cos2(x/y) [cos (x/y)] '
= 3cos2(x/y) [– sin(x/y)] (x/y) '
= 3cos2(x/y) [– sin(x/y)] ((y – xy')/y2)
exy x2+1
= y2 find y' implicitly.
Example C. Given

47. Implicit Equations and Implicit Differentiation
Your turn: Expand [cos3 (x/y)] '
By the Power Chain Rule,
[cos3 (x/y)] ' = 3cos2(x/y) [cos (x/y)] '
= 3cos2(x/y) [– sin(x/y)] (x/y) '
= 3cos2(x/y) [– sin(x/y)] ((y – xy')/y2)
exy x2+1
= y2 find y' implicitly.
Example C. Given
We are to differentiate with respect to x on both sides
of the equation.

48. Implicit Equations and Implicit Differentiation
Your turn: Expand [cos3 (x/y)] '
By the Power Chain Rule,
[cos3 (x/y)] ' = 3cos2(x/y) [cos (x/y)] '
= 3cos2(x/y) [– sin(x/y)] (x/y) '
= 3cos2(x/y) [– sin(x/y)] ((y – xy')/y2)
exy x2+1
= y2 find y' implicitly.
Example C. Given
We are to differentiate with respect to x on both sides
of the equation. However it’s easier if we clear the
denominators first.
y2 exy = x2+1

49. Implicit Equations and Implicit Differentiation
Your turn: Expand [cos3 (x/y)] '
By the Power Chain Rule,
[cos3 (x/y)] ' = 3cos2(x/y) [cos (x/y)] '
= 3cos2(x/y) [– sin(x/y)] (x/y) '
= 3cos2(x/y) [– sin(x/y)] ((y – xy')/y2)
exy x2+1
= y2 find y' implicitly.
Example C. Given
We are to differentiate with respect to x on both sides
of the equation. However it’s easier if we clear the
denominators first.
Hence we differentiate both sides of
y2 exy = x2+1

50. Implicit Equations and Implicit Differentiation
[y2 exy ] ' = [x2+1] '

51. Implicit Equations and Implicit Differentiation
[y2 exy ] ' = [x2+1] '
Product
Rule
(y2)' exy + y2(exy)' =

52. Implicit Equations and Implicit Differentiation
[y2 exy ] ' = [x2+1] '
Product
Rule Derivative of x2 with respect to x
(y2)' exy + y2(exy)' = 2x

53. Implicit Equations and Implicit Differentiation
[y2 exy ] ' = [x2+1] '
Product
Rule Derivative of x2 with respect to x
(y2)' exy + y2(exy)' = 2x
Chain Rule
2y(y)' exy

54. Implicit Equations and Implicit Differentiation
[y2 exy ] ' = [x2+1] '
Product
Rule Derivative of x2 with respect to x
(y2)' exy + y2(exy)' = 2x
Chain Rule
2y(y)' exy + y2 exy (xy)' = 2x

55. Implicit Equations and Implicit Differentiation
[y2 exy ] ' = [x2+1] '
Product
Rule Derivative of x2 with respect to x
(y2)' exy + y2(exy)' = 2x
Chain Rule
2y(y)' exy + y2 exy (xy)' = 2x
Product Chain Rule
2y(y)' exy + y2 exy (y + xy') = 2x

56. Implicit Equations and Implicit Differentiation
[y2 exy ] ' = [x2+1] '
Product
Rule Derivative of x2 with respect to x
(y2)' exy + y2(exy)' = 2x
Chain Rule
2y(y)' exy + y2 exy (xy)' = 2x
Product Chain Rule
2y(y)' exy + y2 exy (y + xy') = 2x
We are done with differentiating.
Next is to extract the y'.

57. Implicit Equations and Implicit Differentiation
[y2 exy ] ' = [x2+1] '
Product
Rule Derivative of x2 with respect to x
(y2)' exy + y2(exy)' = 2x
Chain Rule
2y(y)' exy + y2 exy (xy)' = 2x
Product Chain Rule
2y(y)' exy + y2 exy (y + xy') = 2x
We are done with differentiating.
Next is to extract the y'.
Pull out the inherited exy factor on the left first.
exy [2y(y)' + y2 (y + xy')] = 2x

58. Implicit Equations and Implicit Differentiation
exy [2y(y)' + y2 (y + xy')] = 2x

59. Implicit Equations and Implicit Differentiation
exy [2y(y)' + y2 (y + xy')] = 2x
2y(y)' + y2 (y + xy') = 2x
exy

60. Implicit Equations and Implicit Differentiation
exy [2y(y)' + y2 (y + xy')] = 2x
2y(y)' + y2 (y + xy') = 2x
exy
2y y' + y3 + xy2 y' = 2xexy

61. Implicit Equations and Implicit Differentiation
exy [2y(y)' + y2 (y + xy')] = 2x
2y(y)' + y2 (y + xy') = 2x
exy
2y y' + y3 + xy2 y' = 2xexy
2yy' + xy2 y' = 2x – y3
exy

62. Implicit Equations and Implicit Differentiation
exy [2y(y)' + y2 (y + xy')] = 2x
2y(y)' + y2 (y + xy') = 2x
exy
2y y' + y3 + xy2 y' = 2xexy
2yy' + xy2 y' = 2x – y3
exy
y' (2y+ xy2 ) = 2x – y3
exy

63. Implicit Equations and Implicit Differentiation
exy [2y(y)' + y2 (y + xy')] = 2x
2y(y)' + y2 (y + xy') = 2x
exy
2y y' + y3 + xy2 y' = 2xexy
2yy' + xy2 y' = 2x – y3
exy
y' (2y+ xy2 ) = 2x – y3
exy
2x
exy – y3
y' =
(2y+ xy2 )

64. Implicit Equations and Implicit Differentiation
exy [2y(y)' + y2 (y + xy')] = 2x
2y(y)' + y2 (y + xy') = 2x
exy
2y y' + y3 + xy2 y' = 2xexy
2yy' + xy2 y' = 2x – y3
exy
y' (2y+ xy2 ) = 2x – y3
exy
2x
exy – y3
y' = multiply exy up and down, we get
(2y+ xy2 )
y' = 2x – y3exy
(2y+ xy2 )exy

65. Log Differentiation
The technique of log–differentiation takes advantage
of the fact that a “messy” function or an equation may
be disassembled into simpler terms after logging.

66. Log Differentiation
The technique of log–differentiation takes advantage
of the fact that a “messy” function or an equation may
be disassembled into simpler terms after logging.
To use log–differentiation to find derivative y',

67. Log Differentiation
The technique of log–differentiation takes advantage
of the fact that a “messy” function or an equation may
be disassembled into simpler terms after logging.
To use log–differentiation to find derivative y',
i. takes the log of the function or equation or both
sides and disassemble the formulas as much as
possible in terms of In(#)’s.

68. Log Differentiation
The technique of log–differentiation takes advantage
of the fact that a “messy” function or an equation may
be disassembled into simpler terms after logging.
To use log–differentiation to find derivative y',
i. takes the log of the function or equation or both
sides and disassemble the formulas as much as
possible in terms of In(#)’s.
ii. takes the derivative of the disassembled function
or equation implicitly.

69. Log Differentiation
The technique of log–differentiation takes advantage
of the fact that a “messy” function or an equation may
be disassembled into simpler terms after logging.
To use log–differentiation to find derivative y',
i. takes the log of the function or equation or both
sides and disassemble the formulas as much as
possible in terms of In(#)’s.
ii. takes the derivative of the disassembled function
or equation implicitly.
iii. then collect and isolate the y'.

70. Log Differentiation
The technique of log–differentiation takes advantage
of the fact that a “messy” function or an equation may
be disassembled into simpler terms after logging.
To use log–differentiation to find derivative y',
i. takes the log of the function or equation or both
sides and disassemble the formulas as much as
possible in terms of In(#)’s.
ii. takes the derivative of the disassembled function
or equation implicitly.
iii. then collect and isolate the y'.
Example C.
Use log differentiation to find y' given that
3
2(x)√x2+1
y = sin
e2x– 1

71. Log Differentiation
Takes the natural–log on both sides we have
3
2(x) √x2+1
ln(y) = ln sin
e2x– 1

72. Log Differentiation
Takes the natural–log on both sides we have
3
2(x) √x2+1
ln(y) = ln sin
e2x– 1
ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1)

73. Log Differentiation
Takes the natural–log on both sides we have
3
2(x) √x2+1
ln(y) = ln sin
e2x– 1
ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1)
ln(y) = 2In(sin (x)) + 3 In(x2+1) – 2x + 1
2

74. Log Differentiation
Takes the natural–log on both sides we have
3
2(x) √x2+1
ln(y) = ln sin
e2x– 1
ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1)
ln(y) = 2In(sin (x)) + 3 In(x2+1) – 2x + 1 Differentiate
2
both sides

75. Log Differentiation
Takes the natural–log on both sides we have
3
2(x) √x2+1
ln(y) = ln sin
e2x– 1
ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1)
ln(y) = 2In(sin (x)) + 3 In(x2+1) – 2x + 1 Differentiate
2
both sides
y'
y =

76. Log Differentiation
Takes the natural–log on both sides we have
3
2(x) √x2+1
ln(y) = ln sin
e2x– 1
ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1)
ln(y) = 2In(sin (x)) + 3 In(x2+1) – 2x + 1 Differentiate
2
both sides
y' 2cos(x)
y = sin(x)

77. Log Differentiation
Takes the natural–log on both sides we have
3
2(x) √x2+1
ln(y) = ln sin
e2x– 1
ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1)
ln(y) = 2In(sin (x)) + 3 In(x2+1) – 2x + 1 Differentiate
2
both sides
y' 2cos(x) 3 2x – 2
y = sin(x) + 2 x2+1

78. Log Differentiation
Takes the natural–log on both sides we have
3
2(x) √x2+1
ln(y) = ln sin
e2x– 1
ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1)
ln(y) = 2In(sin (x)) + 3 In(x2+1) – 2x + 1 Differentiate
2
both sides
y' 2cos(x) 3 2x – 2
y = sin(x) + 2 x2+1
Hence
y' = 2cos(x) + 3 2+1
–2 y
sin(x) x

79. Log Differentiation
Takes the natural–log on both sides we have
3
2(x) √x2+1
ln(y) = ln sin
e2x– 1
ln(y) = In(sin2(x)) + In( (x2+1)3/2) – In(e2x–1)
ln(y) = 2In(sin (x)) + 3 In(x2+1) – 2x + 1 Differentiate
2
both sides
y' 2cos(x) 3 2x – 2
y = sin(x) + 2 x2+1
Hence
y' = 2cos(x) + 3 2+1
–2 y
sin(x) x
3
2cos(x) 2(x)√x2+1
y' =
sin(x) + x2+1 – 2 sin
3
e2x– 1