Transcript The Chain Rule

The Chain Rule
Chapter 3.4
The Chain Rule
• The Chain Rule might perhaps be the most important of the
differentiation rules
• It tells us how to differentiation a function within a function
• That is, it tells us how to differentiate a composition of functions
• The next slide shows examples of functions that can be differentiated
without the Chain Rule and those for which the Chain Rule can apply
• Note that for those that use the Chain Rule, each can be thought of as a
composition of functions
The Chain Rule
Without the Chain Rule
𝑦 = 𝑥2 + 1
𝑦 = sin 𝑥
𝑦 = 3𝑥 + 2
𝑦 = 𝑒 𝑥 + tan 𝑥
With the Chain Rule
𝑦=
𝑥2 + 1
𝑦 = sin(6𝑥)
𝑦 = 3𝑥 + 2
5
𝑦 = 𝑒 5𝑥 + tan 𝑥 2
Functions f & g
The Chain Rule
Without the Chain Rule
𝑦 = 𝑥2 + 1
𝑦 = sin 𝑥
𝑦 = 3𝑥 + 2
𝑦 = 𝑒 𝑥 + tan 𝑥
With the Chain Rule
𝑦=
𝑥2 + 1
𝑦 = sin(6𝑥)
𝑦 = 3𝑥 + 2
5
𝑦 = 𝑒 5𝑥 + tan 𝑥 2
Functions f & g
𝑓 𝑥 = 𝑥, 𝑔 𝑥 = 𝑥 2 + 1
The Chain Rule
Without the Chain Rule
𝑦 = 𝑥2 + 1
𝑦 = sin 𝑥
𝑦 = 3𝑥 + 2
𝑦 = 𝑒 𝑥 + tan 𝑥
With the Chain Rule
𝑦=
𝑥2 + 1
𝑦 = sin(6𝑥)
𝑦 = 3𝑥 + 2
Functions f & g
𝑓 𝑥 = 𝑥, 𝑔 𝑥 = 𝑥 2 + 1
𝑓 𝑥 = sin 𝑥 , 𝑔 𝑥 = 6𝑥
5
𝑦 = 𝑒 5𝑥 + tan 𝑥 2
The Chain Rule
Without the Chain Rule
𝑦 = 𝑥2 + 1
𝑦 = sin 𝑥
𝑦 = 3𝑥 + 2
𝑦 = 𝑒 𝑥 + tan 𝑥
With the Chain Rule
𝑦=
𝑥2 + 1
𝑦 = sin(6𝑥)
𝑦 = 3𝑥 + 2
Functions f & g
𝑓 𝑥 = 𝑥, 𝑔 𝑥 = 𝑥 2 + 1
𝑓 𝑥 = sin 𝑥 , 𝑔 𝑥 = 6𝑥
5
𝑦 = 𝑒 5𝑥 + tan 𝑥 2
𝑓 𝑥 = 𝑥 5 , 𝑔 𝑥 = 3𝑥 + 2
The Chain Rule
Without the Chain Rule
𝑦 = 𝑥2 + 1
𝑦 = sin 𝑥
𝑦 = 3𝑥 + 2
𝑦 = 𝑒 𝑥 + tan 𝑥
With the Chain Rule
𝑦=
𝑥2 + 1
𝑦 = sin(6𝑥)
𝑦 = 3𝑥 + 2
Functions f & g
𝑓 𝑥 = 𝑥, 𝑔 𝑥 = 𝑥 2 + 1
𝑓 𝑥 = sin 𝑥 , 𝑔 𝑥 = 6𝑥
5
𝑦 = 𝑒 5𝑥 + tan 𝑥 2
𝑓 𝑥 = 𝑥 5 , 𝑔 𝑥 = 3𝑥 + 2
𝑓 𝑢, 𝑣 = 𝑒 𝑢 + tan 𝑣 , 𝑢 = 5𝑥, 𝑣 = 𝑥 2
Theorem 3.11: The Chain Rule
THEOREM:
If 𝑦 = 𝑓(𝑢) is a differentiable function of u and 𝑢 = 𝑔(𝑥) is a differentiable function of x, then 𝑦 = 𝑓(𝑔 𝑥 ) is
a differentiable function of x and
𝑑𝑦 𝑑𝑢 𝑑𝑦
⋅
=
𝑑𝑢 𝑑𝑥 𝑑𝑥
Or equivalently,
𝑑
𝑓 𝑔 𝑥
= 𝑓 ′ 𝑔 𝑥 ⋅ 𝑔′(𝑥)
𝑑𝑥
PROOF
We will define ℎ 𝑥 = 𝑓(𝑔 𝑥 ) and use the alternative form of the derivative to show that, for 𝑥 = 𝑐
ℎ′ 𝑐 = 𝑓 ′ 𝑔 𝑐 ⋅ 𝑔′(𝑐)
Theorem 3.11: The Chain Rule
THEOREM:
PROOF
First, note that since by assumption, g is differentiable, then it is also continuous. So we know that 𝑔 𝑥 →
𝑔(𝑐) and 𝑥 → 𝑐. We will also assume (the reason will be explained at the end) that there are no other values of x
such that 𝑔 𝑥 = 𝑔(𝑐). Now by definition
ℎ′ 𝑐 = lim
𝑥→𝑐
= lim
𝑥→𝑐
𝑓 𝑔 𝑥
−𝑓 𝑔 𝑐
𝑥−𝑐
𝑓 𝑔 𝑥 −𝑓 𝑔 𝑐
𝑔 𝑥 −𝑔 𝑐
⋅
𝑔 𝑥 −𝑔 𝑐
𝑥−𝑐
Here, we multiplied and divided by 𝑔 𝑥 − 𝑔(𝑐). It is this step that requires us to assume that 𝑔 𝑥 ≠ 𝑔(𝑐) for
any values of x different from c.
Theorem 3.11: The Chain Rule
THEOREM:
PROOF
Continuing
ℎ′
𝑓 𝑔 𝑥 −𝑓 𝑔 𝑐
𝑐 = lim
𝑥→𝑐
𝑔 𝑥 −𝑔 𝑐
𝑓 𝑔 𝑥 −𝑓 𝑔 𝑐
𝑥→𝑐
𝑔 𝑥 −𝑔 𝑐
= lim
= 𝑓′ 𝑔 𝑐
𝑔 𝑥 −𝑔 𝑐
⋅
𝑥−𝑐
𝑔 𝑥 −𝑔 𝑐
𝑥→𝑐
𝑥−𝑐
⋅ lim
⋅ 𝑔′(𝑐)
Theorem 3.11: The Chain Rule
THEOREM:
Although this proof seems reasonable, the problem is that we multiplied and divided by 𝑔 𝑥 − 𝑔(𝑐). This
itself isn’t a problem if we are letting 𝑥 → 𝑐. But if there happen to be other values of x nearby such that
𝑔 𝑥 = 𝑔(𝑐) (as we might find in a function that is not one-to-one), then we will have multiplied by the
0
indefinite form 0. To avoid this, we could define two piecewise functions so that approach to c is continuous.
Appendix A shows how this would work.
The Chain Rule
• When applying the Chain Rule, it is important to identify the “inner”
and the “outer” functions
• If 𝑦 = 𝑓(𝑔 𝑥 ), then g is the “inner” function and f is the “outer”
function
• We may designate a function 𝑢 = 𝑔(𝑥) for that 𝑦 = 𝑓 𝑔 𝑥 = 𝑓(𝑢)
𝑑𝑦
𝑑𝑥
• The Chain Rule says that is the derivative of the “outer” function
(as though 𝑔(𝑥) were just a variable, say u) times the derivative of the
“inner” function
• That is,
𝑑𝑦
𝑑𝑥
= 𝑓 ′ 𝑢 ⋅ 𝑔′ 𝑥 = 𝑓 ′ 𝑔 𝑥
⋅ 𝑔′(𝑥)
Example 2: Decomposition of a Composite
Function
𝒚 = 𝒇(𝒈 𝒙 )
1
𝑦=
𝑥+1
𝑦 = sin(2𝑥)
𝑦=
𝒖 = 𝒈(𝒙)
3𝑥 2 − 𝑥 + 1
𝑦 = tan2 𝑥 = tan 𝑥
2
𝒚 = 𝒇(𝒖)
Example 2: Decomposition of a Composite
Function
𝒚 = 𝒇(𝒈 𝒙 )
1
𝑦=
𝑥+1
𝑦 = sin(2𝑥)
𝑦=
𝒖 = 𝒈(𝒙)
𝑢 =𝑥+1
3𝑥 2 − 𝑥 + 1
𝑦 = tan2 𝑥 = tan 𝑥
2
𝒚 = 𝒇(𝒖)
1
𝑦=
𝑢
Example 2: Decomposition of a Composite
Function
𝒚 = 𝒇(𝒈 𝒙 )
1
𝑦=
𝑥+1
𝑦 = sin(2𝑥)
𝑦=
𝒖 = 𝒈(𝒙)
𝑢 =𝑥+1
𝑢 = 2𝑥
3𝑥 2 − 𝑥 + 1
𝑦 = tan2 𝑥 = tan 𝑥
2
𝒚 = 𝒇(𝒖)
1
𝑦=
𝑢
𝑦 = sin 𝑢
Example 2: Decomposition of a Composite
Function
𝒚 = 𝒇(𝒈 𝒙 )
1
𝑦=
𝑥+1
𝑦 = sin(2𝑥)
𝑦=
𝒖 = 𝒈(𝒙)
𝑢 = 3𝑥 2 − 𝑥 + 1
𝑦= 𝑢
𝑢 =𝑥+1
3𝑥 2 − 𝑥 + 1
𝑦 = tan2 𝑥 = tan 𝑥
𝑢 = 2𝑥
𝒚 = 𝒇(𝒖)
1
𝑦=
𝑢
𝑦 = sin 𝑢
2
Example 2: Decomposition of a Composite
Function
𝒚 = 𝒇(𝒈 𝒙 )
1
𝑦=
𝑥+1
𝑦 = sin(2𝑥)
𝑦=
𝒖 = 𝒈(𝒙)
𝑢 = 3𝑥 2 − 𝑥 + 1
𝑦= 𝑢
𝑢 = tan 𝑥
𝑦 = 𝑢2
𝑢 =𝑥+1
3𝑥 2 − 𝑥 + 1
𝑦 = tan2 𝑥 = tan 𝑥
𝑢 = 2𝑥
𝒚 = 𝒇(𝒖)
1
𝑦=
𝑢
𝑦 = sin 𝑢
2
The Chain Rule Using
𝑑𝑦
𝑑𝑥
Notation
• If we take 𝑦 = 𝑓(𝑔 𝑥 ) and we take 𝑢 = 𝑔(𝑥), then 𝑦 = 𝑓(𝑢)
• In the last version, y is a function of u, so taking the derivative gives
𝑑𝑦
′
𝑓 𝑢 =
𝑑𝑢
𝑑𝑢
′
• However, u is a function of x, so its derivative is 𝑔 𝑥 =
𝑑𝑥
• We combine these to come up with the Chain Rule as follows
𝑑𝑦
𝑑𝑦 𝑑𝑢
′
′
=𝑓 𝑢 ⋅𝑔 𝑥 =
⋅
𝑑𝑥
𝑑𝑢 𝑑𝑥
The Chain Rule Using
•
•
•
•
•
𝑑𝑦
𝑑𝑥
Notation
𝑑𝑦
𝑑𝑦 𝑑𝑢
′
′
=𝑓 𝑢 ⋅𝑔 𝑥 =
⋅
𝑑𝑥
𝑑𝑢 𝑑𝑥
Note that this looks like cross-canceling of du, but it really isn’t
The reason is that derivative is not a ratio, it is a limit
However, under the right circumstances it behaves as though it were a ratio
So, it’s ok to think about cross-canceling here, but do not assume without
justification that these will behave just like ratios in any circumstance
For example, we should not jump to the conclusion that we can add
𝑑𝑦 𝑑𝑢
+
𝑑𝑥 𝑑𝑥
Example 3: Using the Chain Rule
Find
𝑑𝑦
𝑑𝑥
for 𝑦 = 𝑥 2 + 1 3 .
𝑑𝑦
𝑑𝑢
Our inner function here is 𝑢 = 𝑥 2 + 1 while the outer function is 𝑦 = 𝑢3 . Find 𝑑𝑢 and 𝑑𝑥 and multiply them to
get the answer:
𝑑𝑦
𝑑 2
=
𝑥 + 1 = 2𝑥
𝑑𝑢 𝑑𝑢
𝑑𝑢
𝑑 3
=
𝑢 = 3𝑢2
𝑑𝑥 𝑑𝑥
𝑑𝑦 𝑑𝑦 𝑑𝑢
=
⋅
= 3𝑢2 ⋅ 2𝑥 = 3 𝑥 2 + 1
𝑑𝑥 𝑑𝑢 𝑑𝑥
Don’t forget to back-substitute for u!
2
⋅ 2𝑥 = 6𝑥 𝑥 2 + 1
2
Example 3: Using the Chain Rule
Find
𝑑𝑦
𝑑𝑥
for 𝑦 = 𝑥 2 + 1 3 .
For this example we also could have applied the exponent and then taken the derivative (remember that this isn’t
always possible and even when it is, it usually involves more work).
𝑦 = 𝑥2 + 1
So
The result is the same.
3
= 𝑥 6 + 3𝑥 4 + 3𝑥 2 + 1
𝑑𝑦
= 6𝑥 5 + 12𝑥 3 + 6𝑥 = 6𝑥 𝑥 4 + 2𝑥 2 + 1 = 6𝑥 𝑥 2 + 1
𝑑𝑥
2
The General Power Rule
• In an earlier section of this chapter you saw that the Power Rule is
𝑑 𝑛
𝑥 = 𝑛𝑥 𝑛−1
𝑑𝑥
• Using the Chain Rule, we can derive a more general Power Rule that
applies whenever a function is raised to a power
Theorem 3.12: The General Power Rule
THEOREM:
If 𝑦 = 𝑢 𝑥
𝑛,
Or equivalently
where u is a differentiable function of x and n is a rational number, then
𝑑𝑦
𝑑𝑢
𝑛−1
=𝑛 𝑢 𝑥
⋅
𝑑𝑥
𝑑𝑥
𝑑
𝑢 𝑥 𝑛 = 𝑛𝑢𝑛−1 ⋅ 𝑢′
𝑑𝑥
The proof is a straightforward application of the Chain Rule
Example 4: Applying the General Power Rule
Find the derivative of 𝑓 𝑥 = 3𝑥 − 2𝑥 2 3 .
Take 𝑢 = 3𝑥 − 2𝑥 2 . Then we have 𝑓 𝑢 = 𝑢3 and the derivative is
𝑓 ′ 𝑥 = 𝑓 ′ 𝑢 ⋅ 𝑢′ 𝑥 = 3𝑢2 3 − 4𝑥 = 3 3𝑥 − 2𝑥 2 2 (3 − 4𝑥)
Example 5: Differentiating Functions
Involving Radicals
Find all points on the graph of 𝑓 𝑥 =
0.
3
𝑥2 − 1
2
for which 𝑓 ′ 𝑥 =
2
2
3
It will be easier here to rewrite the function with a rational exponent: 𝑓 𝑥 = 𝑥 − 1 . Now take 𝑢 = 𝑥 2 − 1
2
2 −1
2
′
3
and 𝑓 𝑢 = 𝑢 . We have 𝑓 𝑢 = 3 𝑢 3 = 1 and 𝑢′ 𝑥 = 2𝑥. Applying the Chain Rule
3𝑢3
2
1
3𝑢 3
4𝑥
⋅ 2𝑥 =
3
𝑥2
−1
1
3
=
4𝑥
3
3 𝑥2 − 1
Note that if 𝑓 ′ 𝑥 = 0, then the numerator must be zero implying that 𝑥 = 0. Hence, the point at which the
slope of the tangent line is zero is (0,1) (i.e., the value of 𝑓(0)). Confirm this result using your graphing
calculator.
Example 6: Differentiating Quotients With
Constant Numerators
Differentiate 𝑔 𝑡 =
−7
.
2𝑡−3 2
Rewrite the function as 𝑔 𝑡 = −7 2𝑡 − 3
−2
. Now, 𝑢 = 2𝑡 − 3 and 𝑔 𝑢 = −7𝑢−2 . So we have
𝑔′ 𝑢 = 14𝑢−3 and 𝑢′ 𝑡 = 2
So the derivative is
𝑔′ 𝑡 = 𝑔′ 𝑢 ⋅ 𝑢′ 𝑥 = 14𝑢 −3 ⋅ 2 =
28
2𝑡 − 3
3
Simplifying Derivatives
• Finding derivatives of composite functions that also involve products
or quotients requires keeping track of the various rules and where they
must be applied in the function
• It is usually best in these cases to break the derivative into its
component parts and take derivatives separately as needed
• It is best not to skimp on the use of notation
Example 7: Simplifying by Factoring Out the
Least Powers
Find the derivative of 𝑓 𝑥 = 𝑥 2 1 − 𝑥 2 .
First identify the operations in the function. Here we have a product and a radical which can be the “outer”
function in a composite function. Begin by rewriting the radical with a rational exponent
2
𝑓 𝑥 =𝑥 1−
1
2 2
𝑥
Apply the Product rule and use derivative notation for the composite part of the function
1
1
𝑑
′
2
2
2
2
𝑓 𝑥 =𝑥 ⋅
1−𝑥
+ 2𝑥 1 − 𝑥 2
𝑑𝑥
Now let 𝑢 = 1 − 𝑥 2 and 𝑦 = 𝑥 and apply the Chain Rule
1
𝑑
𝑑 1 𝑑
1 −1
1
2
2
2
2
2
1−𝑥
=
𝑢 ⋅
1 − 𝑥 = 𝑢 ⋅ −2𝑥 =
⋅ −2𝑥
𝑑𝑥
𝑑𝑢
𝑑𝑥
2
2 1 − 𝑥2
Replace this in the original derivative and simplify
𝑓′ 𝑥 = 𝑥2 ⋅
−𝑥
1 − 𝑥2
+ 2𝑥 1 − 𝑥 2
Example 7: Simplifying by Factoring Out the
Least Powers
Find the derivative of 𝑓 𝑥 = 𝑥 2 1 − 𝑥 2 .
−𝑥
𝑓′ 𝑥 = 𝑥2 ⋅
=
=
−𝑥 3
1 − 𝑥2
−𝑥 3 + 2𝑥 1 − 𝑥 2
1−
𝑥2
1 − 𝑥2
+ 2𝑥 1 − 𝑥 2
+ 2𝑥 1 − 𝑥 2 ⋅
=
1 − 𝑥2
1 − 𝑥2
−𝑥 3 + 2𝑥 − 2𝑥 3
1−
𝑥2
=
2𝑥 − 3𝑥 3
1 − 𝑥2
Example 8: Simplifying the Derivative of a
Quotient
Find the derivative of 𝑓 𝑥 =
𝑥
3
𝑥 2 +4
Rewrite the denominator with a rational exponent
𝑓 𝑥 =
𝑥
𝑥2
Apply the Quotient Rule
𝑓′
𝑥 =
𝑥2
+4
1
3
+4
1
3
𝑑 2
⋅1−𝑥⋅
𝑥 +4
𝑑𝑥
𝑥2 + 4
1
3
1
3
2
Apply the Chain Rule with 𝑢 = 𝑥 2 + 4
1
𝑑 2
𝑑 1 𝑑 2
1 −2
𝑥 +4 3 =
𝑢3 ⋅
𝑥 + 4 = 𝑢 3 ⋅ 2𝑥 =
𝑑𝑥
𝑑𝑢
𝑑𝑥
3
2𝑥
3
𝑥2
+4
2
3
Example 8: Simplifying the Derivative of a
Quotient
Find the derivative of 𝑓 𝑥 =
𝑥
3
𝑥 2 +4
Replace this in the original derivative and simplify
𝑥2
𝑓′
+4
1
3
2𝑥
−𝑥⋅
3
𝑥 =
𝑥2
+4
𝑥2
+4
2
3
2
3
3(𝑥 2 + 4) − 2𝑥 2
=
2
3
+4 3
2
2
𝑥 +4 3
𝑥2
𝑥 2 + 12
=
3
𝑥2
+4
4
3
=
3𝑥 2 + 12 − 2𝑥 2
3
𝑥2
+4
4
3
Example 9: Simplifying the Derivative of a
Power
Find the derivative of 𝑦 =
3𝑥−1 2
𝑥 2 +3
We can apply the General Power Rule
𝑦′ = 2
3𝑥 − 1
𝑑 3𝑥 − 1
⋅
𝑥 2 + 3 𝑑𝑥 𝑥 2 + 3
Use the Quotient Rule
𝑑 3𝑥 − 1
𝑥 2 + 3 ⋅ 3 − 3𝑥 − 1 ⋅ 2𝑥 3𝑥 2 + 9 − 6𝑥 2 + 2𝑥 −3𝑥 2 + 2𝑥 + 9
=
=
=
𝑑𝑥 𝑥 2 + 3
𝑥2 + 3 2
𝑥2 + 3 2
𝑥2 + 3 2
Substitute this into the original derivative and simplify
𝑦′
2 3𝑥 − 1 −3𝑥 2 + 2𝑥 + 9 2 3𝑥 − 1 −3𝑥 2 + 2𝑥 + 9
=
⋅
=
𝑥2 + 3
𝑥2 + 3 2
𝑥2 + 3 3
Trigonometric Functions & the Chain Rule
• The following show the application of the Chain Rule to trigonometric
functions
• Remember that these apply when the argument is itself a function
• That is, in the table that follows, u will be a function of x
Trigonometric Functions & the Chain Rule
𝒅
sin 𝒖 = 𝒖′ cos 𝒖
𝒅𝒙
𝒅
cos 𝒖 = −𝒖′ sin 𝒖
𝒅𝒙
𝒅
tan 𝒖 = 𝒖′ sec 2 𝒖
𝒅𝒙
𝒅
csc 𝒖 = −𝒖′ csc 𝒖 cot 𝒖
𝒅𝒙
𝒅
sec 𝒖 = 𝒖′ sec 𝒖 tan 𝒖
𝒅𝒙
𝒅
cot 𝒖 = −𝒖′ csc 2 𝒖
𝒅𝒙
𝒅 𝒖
𝒆 = 𝒖′ 𝒆𝒖
𝒅𝒙
Example 10: Applying the Chain Rule to
Transcendental Functions
a) 𝑦 = sin 2𝑥, with 𝑢 = 2𝑥. 𝑦 ′ = 𝑢′ cos 𝑢 = 2 cos 2𝑥
b) 𝑦 = cos(𝑥 − 1), with 𝑢 = 𝑥 − 1. 𝑦 ′ = −𝑢′ sin 𝑢 = − 1 sin 𝑥 − 1 = − sin(𝑥 − 1)
c) 𝑦 = 𝑒 3𝑥 , with 𝑢 = 3𝑥. 𝑦 ′ = 𝑢′ 𝑒 𝑢 = 3𝑒 3𝑥
Example 11: Parentheses & Trigonometric
Functions
a) 𝑦 = cos 3𝑥 2 = cos 3𝑥 2 , with 𝑢 = 3𝑥 2 . 𝑦 ′ = −𝑢′ sin 𝑢 = −6𝑥 sin 3𝑥 2
b) 𝑦 = cos 3 𝑥 2 . Recognize that this is just a constant times 𝑥 2 , so the Chain Rule does not apply. We get
𝑦 ′ = 2𝑥 cos 3
c) 𝑦 = cos 3𝑥
2
= cos 9𝑥 2 , with 𝑢 = 9𝑥 2 . 𝑦 ′ = −𝑢′ sin 𝑢 = −18𝑥 sin 9𝑥 2
d) y = cos 2 𝑥 = cos 𝑥 2 . Apply the General Power Rule: 𝑦 ′ = 2 cos 𝑥 ⋅ − sin 𝑥 = −2 sin 𝑥 cos 𝑥
Repeated Application of the Chain Rule
• A function of the form 𝑘 𝑥 = 𝑓 𝑔 ℎ 𝑥
requires two applications
of the chain rule
• To see why, let 𝑢 = ℎ(𝑥), 𝑣 = 𝑔(𝑢), and 𝑦 = 𝑓(𝑣)
• By the Chain Rule we get
𝑑𝑦 𝑑𝑦 𝑑𝑣 𝑑𝑢
=
⋅
⋅
𝑑𝑥 𝑑𝑣 𝑑𝑢 𝑑𝑥
• In the next example, use u and v for the inner functions
Example 12: Repeated Application of the
Chain Rule
Find the derivative of 𝑓 𝑡 = sin3 4𝑡
First, rewrite the function as 𝑓 𝑡 = sin 4𝑡 3 . Now take 𝑢 = 4𝑡, 𝑣 = sin 𝑢, and 𝑦 = 𝑣 3 . Differentiate each of
these
𝑑𝑦
𝑑𝑣
𝑑𝑢
2
= 3𝑣 ,
= cos 𝑢 ,
=4
𝑑𝑣
𝑑𝑢
𝑑𝑡
𝑑𝑦
Remember that 𝑑𝑥 =
𝑑𝑦
𝑑𝑣
𝑑𝑣
𝑑𝑢
⋅ 𝑑𝑢 ⋅ 𝑑𝑡
𝑑𝑦
= 3𝑣 2 ⋅ cos 𝑢 ⋅ 4 = 12𝑣 2 cos 4𝑡 = 12 sin2 𝑢 cos 4𝑡 = 12 sin2 4𝑡 cos 4𝑡
𝑑𝑥
Derivative of the Natural Logarithm Function
• Up to this point, the derivatives of algebraic functions (like
polynomial functions or radical functions) have been algebraic
functions
• Likewise, the derivatives of transcendental functions (like
trigonometric functions and exponential functions) have been
transcendental functions
• The natural logarithm function is unusual in that it is a transcendental
function, but its derivative is an algebraic function
Theorem 3.13: Derivative of the Natural
Logarithm Function
THEOREM:
Let u be a differentiable function of x. Then
1.
𝑑
𝑑𝑥
2.
𝑑
𝑑𝑥
1
ln 𝑥 = 𝑥 , 𝑥 > 0
ln 𝑢 =
𝑢′
,𝑢
𝑢
>0
PROOF
Rewrite the function 𝑦 = ln 𝑥 as 𝑒 𝑦 = 𝑥. Now differentiate both sides of the equation with respect to x
𝑑 𝑦
𝑑
𝑒 =
[𝑥]
𝑑𝑥
𝑑𝑥
𝑒𝑦
𝑑𝑦
=1
𝑑𝑥
Theorem 3.13: Derivative of the Natural
Logarithm Function
THEOREM:
𝑑𝑦
=1
𝑑𝑥
Note here that we have applied the Chain Rule to obtain the result on the left because y is a function of x.
Another way to think of this is to set 𝑦 = 𝑓(𝑥) and we get
𝑑 𝑓𝑥
𝑑𝑦
𝑓
𝑥
′
𝑦
𝑒
=𝑒
⋅𝑓 𝑥 =𝑒
𝑑𝑥
𝑑𝑥
𝑒𝑦
𝑑𝑦
Solve for 𝑑𝑥
𝑑𝑦
1
1
=
=
𝑑𝑥 𝑒 𝑦 𝑥
Example 13: Differentiation of the
Logarithmic Functions
a) 𝑦 = ln 2𝑥, with 𝑢 = 2𝑥.
𝑑
𝑑𝑥
𝑑
𝑑
1
2
1
ln 2𝑥 = 𝑑𝑢 ln 𝑢 ⋅ 𝑑𝑥 2𝑥 = 𝑢 ⋅ 2 = 2𝑥 = 𝑥
b) 𝑦 = ln(𝑥 2 + 1), with 𝑢 = 𝑥 2 + 1.
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑢
1
1
2𝑥
= 𝑑𝑢 ⋅ 𝑑𝑥 = 𝑢 ⋅ 2𝑥 = 𝑥 2+1 ⋅ 2𝑥 = 𝑥 2+1
𝑑
𝑑
1
c) 𝑓 𝑥 = 𝑥 ln 𝑥. Apply the Product Rule: 𝑓 ′ 𝑥 = 𝑥 ⋅ 𝑑𝑥 ln 𝑥 + ln 𝑥 ⋅ 𝑑𝑥 𝑥 = 𝑥 ⋅ 𝑥 + 1 ⋅ ln 𝑥 = 1 + ln 𝑥
1
d) 𝑓 𝑥 = ln 𝑥 3 , with 𝑢 = ln 𝑥. 𝑓 ′ 𝑥 = 𝑓 ′ 𝑢 ⋅ 𝑢′ 𝑥 = 3𝑢2 ⋅ 𝑥 = 3 ln 𝑥
2
1
⋅𝑥 =
3 ln 𝑥 2
𝑥
Logarithmic Properties as Aids to
Differentiation
• Logarithms were first developed by amateur mathematician John
Napier during the 16th century as an aid to calculation, specifically
multiplying, dividing, and raising to powers
• Prior to the widespread availability of calculators, doing such
calculations was tedious (as they still are)
• Napier created tables of logarithmic values so that products and
quotients could be turned into sums and differences using the
properties of logarithms
• Though such tables are now obsolete, we can use the logarithm
properties to simplify differentiation, as the following examples show
Example 14: Logarithmic Properties as Aids
to Differentiation
Differentiate 𝑓 𝑥 = ln 𝑥 + 1
We could apply the Chain Rule twice here, but we can more easily make use of the logarithm properties by
rewriting
1
1
𝑓 𝑥 = ln 𝑥 + 1 = ln 𝑥 + 1 2 = ln 𝑥 + 1
2
Now,
1 𝑑
1
1
1
𝑓′ 𝑥 = ⋅
ln 𝑥 + 1 =
=
2 𝑑𝑥
2 𝑥+1
2 𝑥+1
Example 15: Logarithmic Properties as Aids
to Differentiation
Differentiate 𝑓 𝑥 = ln
𝑥
2
2
𝑥 +1
2𝑥 3 −1
Again we could apply the Chain Rule here, but using the logarithm properties greatly simplifies the task
𝑓 𝑥 = ln 𝑥
𝑥2
+1
2
2𝑥 3
−1
−1
2
1
= ln 𝑥 + 2 ln 𝑥 2 + 1 − ln(2𝑥 3 − 1)
2
Differentiating
2
1
1
1
1
1
4𝑥
3𝑥
𝑓 ′ 𝑥 = + 2 ⋅ 2𝑥 ⋅ 2
− ⋅ 6𝑥 2 ⋅ 3
= +
−
𝑥
𝑥 +1 2
2𝑥 − 1 𝑥 𝑥 2 + 1 2𝑥 3 − 1
Derivative of the Natural Logarithm
Involving Absolute Value
• Since logarithms are undefined for 𝑥 < 0, it is common to see the
form 𝑦 = ln |𝑥|
• As the next theorem shows, we can differentiate such forms as though
the absolute value signs were missing
Theorem 3.14: Derivative Involving Absolute
Value
THEOREM:
If u is a differentiable function of x such that 𝑢 ≠ 0, then
𝑑
𝑢′
ln 𝑥 =
𝑑𝑥
𝑢
PROOF
If 𝑢 > 0, then 𝑢 = 𝑢 and the result is the same as Theorem 3.13. If 𝑢 < 0, then 𝑢 = −𝑢 and we have
𝑑
𝑢′
𝑢′
𝑑
ln −𝑢 = −
= =
[ln 𝑢]
𝑑𝑥
−𝑢 𝑢 𝑑𝑥
Exponential Function to Base a
DEFINITION:
If a is a positive real number (𝑎 ≠ 1) and x is any real number, then the exponential function to base a is
denoted by 𝑎 𝑥 as is defined by
𝑎 𝑥 = 𝑒 𝑥 ln 𝑎
If 𝑎 = 1, then 𝑦 = 1𝑥 = 1 is a constant function.
Logarithmic Function to Base a
DEFINITION:
If a is a positive real number (𝑎 ≠ 1) and x is any positive real number, then the logarithmic function to the
base a is denoted by log 𝑎 𝑥 and is defined as
1
log 𝑎 𝑥 =
⋅ ln 𝑥
ln 𝑎
Note that this can be thought of as the change of base formula applied to base a with input x.
Theorem 3.15: Derivatives for Bases Other
Than e
THEOREM:
Let a be a positive real number other than 1 and let u be a differentiable function of x.
1.
𝑑
𝑑𝑥
𝑎 𝑥 = 𝑎 𝑥 ln 𝑎
2.
𝑑
𝑑𝑥
𝑎𝑢 = 𝑢′ 𝑎𝑢 ln 𝑢
3.
𝑑
𝑑𝑥
log 𝑎 𝑥 = 𝑥 ln 𝑎
4.
𝑑
𝑑𝑥
𝑢′
𝑢 ln 𝑎
1
log 𝑎 𝑢 =
Example 16: Differentiating Functions to
Other Bases
Find the derivative of each of the following
a) 𝑦 = 2𝑥 , 𝑦 ′ = 2𝑥 ln 2
b) 𝑦 = 23𝑥 , with 𝑢 = 3𝑥.
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑢
= 𝑑𝑢 ⋅ 𝑑𝑥 = 2𝑢 ln 2 ⋅ 3 = 23𝑥 ⋅ 3 ln 2
𝑑𝑦
𝑑𝑢
1
− sin 𝑥
c) 𝑦 = log10 cos 𝑥, with 𝑢 = cos 𝑥. 𝑦 ′ = 𝑑𝑢 ⋅ 𝑑𝑥 = 𝑢 ln 10 ⋅ − sin 𝑥 = cos 𝑥 ln 10 =
− tan 𝑥
ln 10
Exercise 3.4a
• Page 161, #1-36
Exercise 3.4b
• Page 161, #49-99 odds
Exercise 3.4c
• Page 162, #101-137 odds