3. Implicit Differentiation
β’ Most of you are sometimes face with situations where
you need to choose between expressing explicitly and
expressing implicitly. Of course, you will choose the
one that is clear and not ambiguous. The same
situation happens also in mathematics, especially
when dealing with functions. A function is written in
explicit form when it is expressed as π¦ = π(π₯). All
functions that are dealt with in the previous modules
are written in this form.
4. Implicit Differentiation
β’However, there are times when mathematical
functions are written in more complicated form,
wherein it is difficult (or sometimes impossible)
to express π¦ explicitly in terms of π₯ . Such
functions are expressed in implicit form. This
section of the module will try to explain how
these functions can be differentiated.
6. Implicit Differentiation
β’The first function defines π¦ as a function
of π₯ explicitly, because for each π₯, the equation
gives the explicit formula π¦ = π π₯ for finding
the corresponding value of π¦. You can easily
differentiate this function following the usual
process by applying the differentiation rules:
ππ¦
ππ₯
= 2π₯
7. Implicit Differentiation
β’The second function may still be differentiated
easily, provided that it is possible to express it in
explicit form. Because π₯2
+ 3π¦ = 4 , then by
algebraic manipulation, π¦ = β
1
3
π₯2
+
4
3
. You can
now differentiate this function and yield
ππ¦
ππ₯
= β
1
3
2π₯ = β
2
3
π₯
8. Implicit Differentiation
β’However, there are cases when it is totally
impossible to rewrite the given equation as
function of π₯, such as π₯4
+ 2π₯π¦ β π¦3
+ 2 = 0. It
is in these instances that implicit differentiation
is used.
9. Implicit Differentiation
β’ Because the equation is written in implicit form, the
process of implicit differentiation involves
differentiating each term of an equation on both
sides, including those that contain the variably π¦. To
totally understand how this works, remember that
you are always differentiating in terms of π₯. Thus, you
must differentiate the π₯ terms in the usual way. But
for the π¦ terms, the chain rule must be applied
because π¦ is defined implicitly as a function of π₯. For
example, the expression π₯3
+ 3π¦2
has a derivative
of 3π₯2
+ 6π¦
ππ¦
ππ₯
.
10. Implicit Differentiation
β’ To further guide you on the implicit differentiation
process, follow these steps:
β’ Differentiate both sides of the equation with respect
to π₯.
β’ Combine all terms containing
ππ¦
ππ₯
on the left side of the
equation and all the other terms on the right side.
β’ On the left side of the equation, factor out
ππ¦
ππ₯
.
β’ Isolate
ππ¦
ππ₯
by dividing out the other factor on the left side.
12. Solution:
β’Differentiate both sides of the equation with
respect to π₯.
π
ππ₯
π¦3
+ π¦ β 3π₯2
=
π
ππ₯
2π₯ + 5
β’Take derivative of each term with respect
to π₯ and use chain rule.
3π¦2
ππ¦
ππ₯
+
ππ¦
ππ₯
β 6π₯ = 2
13. Solution:
β’Combine all terms with
ππ¦
ππ₯
on the left side.
3π¦2
ππ¦
ππ₯
+
ππ¦
ππ₯
= 6π₯ + 2
β’Factor out
ππ¦
ππ₯
.
ππ¦
ππ₯
3π¦2
+ 1 = 6π₯ + 2
20. Solution:
β’The equation is that of unit circle. Thus, it is not
a function. We can, however, restrict to apply
implicit differentiation.
β’Differentiate both sides with respect to π₯.
π
ππ₯
π₯2
+ π¦2
=
ππ¦
ππ₯
1
π
ππ₯
π₯2
+
π
ππ₯
π¦2
= 0
22. Solution:
β’To find the slope at point at point
1
2
,
3
2
,
substitute
1
2
for π₯ and
3
2
for π¦. You now obtain.
β’
ππ¦
ππ₯
= β
π₯
π¦
= β
1
2
3
2
= β
1
2
β
2
3
= β
1
3
23. Example 4:
β’Find the equation of the tangent line to the
curve π₯3
+ π¦3
= 9 at the point 1, 2 .
24. Solution:
β’Sometimes, you can use π¦β²
instead of
ππ¦
ππ₯
.
Differentiating implicitly with respect to π₯,
β’
ππ¦
ππ₯
π₯3
+ π¦3
=
ππ¦
ππ₯
9
β’3π₯2
+ 3π¦2
π¦β²
= 0
β’3π¦2
π¦β²
= β3π₯2
β’π¦β²
= β
3π₯2
3π¦2 = β
π₯2
π¦2
25. Solution:
β’Now, at the point 1, 2 , the slope π of the
tangent line is π = β
π₯2
π¦2 = β
1 2
2 2 = β
1
4
. Using the
point-slope form of the equation of the line, you
now have
β’π¦ β 2 = β
1
4
π₯ β 1
β’π¦ = β
1
4
π₯ +
1
4
+ 2
β’π¦ = β
1
4
π₯ +
9
4
26. Example 5:
β’ Consider the equation π₯2
+ π¦2
= 9.
a) Find π¦β²
by implicit differentiation.
b) Derive two functions that can be defined
from the equation.
c) Find the derivative of each functions
obtained in (b) by explicit differentiation.
d) Verify that the result obtained in (a) agrees
with result in (c).
32. Big Idea:
β’The implicit differentiation is performed on
functions that are difficult to express directly. As a
result, the differentiation process takes longer
than usual and also a bit confusing when not
properly carried out. This is the same for the
things you cannot express directly. They may imply
the right message, but the receiver might
interpret them incorrectly when properly stated.
34. Derivative Rules for exponential and Logarithmic
Functions with Base π
β’You probably have now mastered how to
differentiate algebraic functions based on the
examples and exercises in the previous module.
Not all functions, however, are algebraic in form,
so you cannot apply the techniques for
differentiation right away. In the case of
exponential and logarithmic functions, they also
have a set of rules that you must follows in
finding the derivatives.
35. Derivative Rules for exponential and Logarithmic
Functions with Base π
β’ Among all the exponential functions of the form π¦ = ππ₯
,
natural exponential function is of great importance in
terms of differentiation. Recall that the natural
exponential function is written as π¦ = ππ₯
, where π is the
Euler number, an irrational number equivalent
to 2.718281828 β¦. The most important thing about the
natural exponential function is that its derivative is
itself. Hence,
π
ππ₯
ππ₯
= ππ₯
. To prove this, examine the
value of π once again. You may want to verify these
values using your calculator.
36. Derivative Rules for exponential and Logarithmic
Functions with Base π
Table 12.1. Values of π as π₯ approaches 0
π₯ 1 + π₯
(1 + π₯)
1
π₯
0.1 1.1 2.59374246
0.01 1.01 2.704813829
0.001 1.001 2.716923932
0.0001 1.0001 2.718145927
0.00001 1.00001 2.718268237
0.000001 1.000001 2.718280469
0.0000001 1.0000001 2.718281694
0.00000001 1.00000001 2.718281786
37. Derivative Rules for exponential and Logarithmic
Functions with Base π
β’From table 12.1, the value of
(1 + π₯)
1
π₯ approaches the
number π as π₯ approaches 0. Thus,
lim
π₯β0
(1 + π₯)
1
π₯= π. This implies that as π₯ becomes
sufficiently small, π = (1 + π₯)
1
π₯ or ππ₯
= 1 + π₯ .
Now, suppose π π₯ = ππ₯
. Then,
38. Derivative Rules for exponential and Logarithmic
Functions with Base π
πβ²(π₯) = lim
ββ0
ππ₯+β
β ππ₯
β
Definition of derivative
= lim
ββ0
ππ₯
(πβ
β 1)
β
Factoring out ππ₯
= lim
ββ0
ππ₯
(1 + β β 1)
β
Based on πβ
= 1 + β
= lim
ββ0
ππ₯
(β)
β
Simplifying terms
= lim
ββ0
ππ₯
= ππ₯
39. Derivative of Natural Exponential Functions
β’The natural exponential functions is its own
derivative, that is,
π
ππ₯
ππ₯
= ππ₯
β’Generally, if π’ is a differentiable function of π₯,
then
π
ππ₯
ππ’
= ππ’
ππ’
40. Example 6:
β’Differentiate each function.
a) π π₯ = π3π₯
b) π π₯ = πβ2π₯2
c) π π₯ = π₯π2π₯
d) β π₯ =
π2π₯+1
ππ₯β1
43. β’Now that you know the derivative of the natural
exponential function, you can now derive the rule
for the derivative of its inverse, which is the
natural logarithmic function. The derivative of the
natural logarithmic function may be derived by
using implicit differentiation. Suppose that π¦ =
ln π₯.
45. Derivative of Natural Logarithmic Function
β’The derivative of the natural logarithmic function
is the reciprocal function, that is,
π
ππ₯
ln π₯ =
1
π₯
β’If π’ is differentiable function of π₯, then
π
ππ₯
ln π’ =
1
π’
ππ’
46. Example 7:
β’Find the derivative of each function.
a) π¦ = ln 4π₯
b) π π₯ = ln(5π₯2
+ 3)
c) π π₯ =
π₯
ln π₯
d) β π₯ = π₯ ln π₯
51. Solution:
β’Recall that the quotient rule of logarithms states
that the logarithm of a quotient is equal to the
difference of the logarithm of the numerator and
the logarithm of the denominator.
π π₯ = ln
4π₯ + 2
π₯ β 1 2
= ln 4π₯ + 2 β ln π₯ β 1 2
β’By power rule of logarithms,
π π₯ = ln 4π₯ + 2 β 2 ln(π₯ β 1)
54. Derivative Rules for General Exponential and
Logarithmic Functions
β’When the given exponential or logarithmic
function expressed using a base other than the
number π, the following rules may be applied.
These rules can be derived through rewriting the
function using base π.
55. Derivative with Other Bases
β’ Let π’ be a differentiable function of π₯ and π is a real
number such that π > 0 and π β 1. Then,
β’
π
ππ₯
ππ₯
= (ln π)ππ₯
β’
π
ππ₯
logπ π₯ =
1
ln π
1
π₯
β’
π
ππ₯
ππ’
= (ln π)ππ’
ππ’
β’
π
ππ₯
logπ π’ =
1
ln π
1
π’
ππ’
61. Example 1:
β’Find the derivative of the following function:
a) π¦ = 52π₯
b) π¦ = ππ₯
+ 8π₯
c) π¦ = log5(3π₯2
+ 1)
d) π¦ = log π₯ π₯ β 2 3
64. Solution:
β’This function will be best differentiated if it will be
simplified first using the product rule for logarithms, i.e.,
log π’π£ = log π’ + log π£. Also, recall that this is a case of
common logarithm in which the base, although not
written, is understood to be 10.
log π₯ π₯ β 2 3
= log π₯ + log π₯ β 2 3
= log π₯ + 3 log(π₯ β 2)
π
ππ₯
log π₯ π₯ β 2 3
=
π
ππ₯
log π₯ + 3 log(π₯ β 2)
67. Derivative Rules for Trigonometric Functions
β’You will deepen your understanding of
trigonometric functions as you explore their
derivatives. The trigonometric functions have
many useful applications in several fields such
as physics, astronomy, and biological sciences.
Needed in the formulation of the rule for the
derivatives of the six trigonometric functions are
the two important trigonometric limits:
lim
π₯β0
sin π₯
π₯
= 1 and lim
π₯β0
1βcos π₯
π₯
= 0.
68. Derivative Rules for Trigonometric Functions
β’The derivations of the differentiation formulas
for trigonometric functions are based on these
two limits. To start, let us derive the formula
for
π
ππ₯
sin π₯ as follows.
β’By the definition of the derivative,
π
ππ₯
sin π₯ = lim
ββ0
sin π₯ + β β sin π₯
β
69. Derivative Rules for Trigonometric Functions
β’Using the sum identity for sine, you have sin π₯ + β =
sin π₯ cos β + cos π₯ sin β. Then,
π
ππ₯
sin π₯ = lim
ββ0
sin π₯ cos β + cos π₯ sin β β sin π₯
β
= lim
ββ0
cos π₯ sin β
β
+
sin π₯ cos β β sin π₯
β
= lim
ββ0
cos π₯ sin β
β
+ lim
ββ0
sin π₯ cos β β sin π₯
β
70. Derivative Rules for Trigonometric Functions
= lim
ββ0
cos π₯ β
sin β
β
+ lim
ββ0
β sin π₯
= cos π₯ lim
ββ0
sin β
β
β sin π₯ lim
ββ0
1 β cos β
β
= cos π₯ 1 β sin π₯ (0)
π
ππ₯
sin π₯ = cos π₯
71. Derivative Rules for Trigonometric Functions
β’Thus, the derivative of the sine function is the cosine
function. To derive the differentiation formula forcos π₯, you
have
π
ππ₯
cos π₯ = lim
ββ0
cos π₯ + β β cos π₯
β
= lim
ββ0
cos π₯ cos β β sin π₯ sin β β cos π₯
β
= lim
ββ0
cos π₯ cos β β cos π₯ β (sin π₯ sin β)
β
72. Derivative Rules for Trigonometric Functions
= lim
ββ0
cos π₯ cos β β cos π₯
β
= β cos π₯ lim
ββ0
1 β cos β
β
= β cos π₯ 0 β sin π₯ (1)
π
ππ₯
cos π₯ = β sin π₯
73. Derivative Rules for Trigonometric Functions
β’In general, for any function π’ differentiable at
any value of π₯ ,
π
ππ₯
sin π’ =
cos π’ ππ’ and
π
ππ₯
cos π’ = β sin π’ ππ’ . The
following gives the differentiation formulas for
the other functions. Their derivatives are left for
you as an exercise.
74. Derivatives of the Trigonometric Functions
β’
π
ππ₯
sin π’ = cos π’ ππ’
β’
π
ππ₯
cos π’ = β sin π’ ππ’
β’
π
ππ₯
tan π’ = sec2
π’ ππ’
82. Solution:
β’ Let π’ = sin π₯ and π£ = 1 β 2 cos π₯ , giving ππ’ =
cos π₯ and ππ£ = 2 sin π₯. Applying quotient rule, you have
πβ²
π₯ =
1 β 2 cos π₯ cos π₯ β sin π₯ 2 sin π₯
1 β 2 cos π₯ 2
=
cos π₯ β 2cos2
π₯ β 2sin2
π₯
1 β 2 cos π₯ 2
83. Solution:
=
cos π₯ β 2(cos2
π₯ + sin2
π₯)
1 β 2 cos π₯ 2
=
cos π₯ β 2(1)
1 β 2 cos π₯ 2
=
cos π₯ β 2
1 β 2 cos π₯ 2
85. Solution:
β’ Apply the quotient rule followed by chain rule. Then
π¦β²
=
4 + 5 cos 2π₯ 3 cos 3π₯ β sin 3π₯ β10 sin 2π₯
4 + 5 cos 2π₯ 2
=
12 cos 3π₯ + 15 cos 2π₯ cos 3π₯ + (10 sin 2π₯ sin 3π₯)
4 + 5 cos 2π₯ 2
=
12 cos 3π₯ + 15 cos 2π₯ cos 3π₯ + 10 sin 2π₯ sin 3π₯
4 + 5 cos 2π₯ 2
87. Solution:
β’ Differentiating implicitly with respect to π₯, you get
π₯ β β sin π¦
ππ¦
ππ₯
+ cos π¦ 1 + π¦ β β sin π₯ + cos π₯
ππ¦
ππ₯
= 0
βπ₯ sin π¦
ππ¦
ππ₯
+ cos π¦ β π¦ sin π₯ + cos π₯
ππ¦
ππ₯
= 0
β sin π¦
ππ¦
ππ₯
+ cos π₯
ππ¦
ππ₯
= π¦ sin π₯ β cos π¦
ππ¦
ππ₯
cos π₯ β π₯ sin π¦ = π¦ sin π₯ β cos π¦
ππ¦
ππ₯
=
π¦ sin π₯ β cos π¦
cos π₯ β π₯ sin π¦
91. Example 7:
β’In a certain city, the normal average
temperature (in degrees Fahrenheit) is modelled
by the function π π‘ = 49 β 18 cos
π(π‘β32)
180
,
where π‘ is the time in days so that π‘ =
1 corresponds to the first of January. What is the
expected date of the warmest day and the
coolest day?
92. Solution:
β’ To solve this problem, you need to find the value of π‘ such
that πβ²
π‘ = 0.
πβ²
π‘ = 0 β 18
π
ππ‘
cos
π π‘ β 32
180
= β18 β sin
π π‘ β 32
180
β
π
ππ‘
π π‘ β 32
180
= 18 sin
π(π‘ β 32)
180
β
π
180
=
π
10
sin
π π‘ β 32
180
97. Solution:
β’ For π = 2π onward, π‘ is already out of the domain because a
year has only 365 days. Substituting n the original function,
if π‘ = 32,
π π‘ = 49 β 18 cos
π 32 β 32
180
= 49 β 18 cos 0 = 49 β 18 1
= 31β
β’ If π‘ = 212,
π π‘ = 49 β 18 cos
π 212 β 32
180
= 49 β 18 cos 0
= 49 β 18 β1
= 67β
98. Solution:
β’ Therefore, the coolest day will happen when π‘ = 32, i.e.,
the first of February, and the warmest day will be
when π‘ = 212, or 31st of July, provided that it is not leap
year.