Unit 2 – Differentiation
Objectives:
1.
2.
3.
4.
Find the derivative of a composite function using the Chain Rule.
Find the derivative of a function using the General Power Rule.
Simplify the derivative of a function using algebra.
Find the derivative of a trigonometric function using the Chain Rule.
The Chain Rule is an extremely important rule in calculus and allows
us to extend differentiation to many more types of functions. Below
are examples of functions that can be done without the Chain Rule,
and those that can be done with it.
Without the Chain Rule
With the Chain Rule
𝑦 = 𝑥2 + 1
𝑦=
𝑦 = sin(𝑥)
𝑦 = sin(6𝑥)
𝑦 = 3𝑥 + 2
𝑦 = 𝑥 + tan(𝑥)
𝑥2 + 1
𝑦 = 3𝑥 + 2
5
𝑦 = 𝑥 + tan(𝑥 2 )
The Chain Rule expands differentiation to composite functions
like 𝑓 ∘ 𝑔 𝑥 = 𝑓 𝑔 𝑥 or 𝑔 ∘ 𝑓 𝑥 = 𝑔 𝑓 𝑥 .
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
𝑑𝑥
=
𝑑𝑦
𝑑𝑢
∙
𝑑𝑢
𝑑𝑥
or, equivalently,
𝑑
𝑑𝑥
𝑓(𝑔 𝑥 ) = 𝑓 ′ 𝑔 𝑥 𝑔′ 𝑥 .
» Part of the process in using the Chain Rule is
figuring out the best way to decompose a
function. Remember that a composite function
is made from one function inside of another.
For example, 𝑦 = 3𝑥 can be thought of as the
function 𝑔(𝑥) = 𝑢 = 3𝑥 inside of the function
𝑓, where 𝑓 𝑢 = 𝑢.
» Example 1 will illustrate this process further.
» Decompose the following functions:
𝑦 = 𝑓(𝑔 𝑥 )
𝑢 = g(𝑥)
𝑦 = 𝑓(𝑢)
𝑢 = 2𝑥 − 3
1
𝑦=
𝑢
B. 𝑦 = cos(3𝑥)
𝑢 = 3𝑥
𝑦 = cos(𝑢)
C. 𝑦 = 5𝑥 2 + 3𝑥
𝑢 = 5𝑥 2 + 3𝑥
𝑦= 𝑢
D. 𝑦 = cot 2 (𝑥)
𝑢 = cot(𝑥)
𝑦 = 𝑢2
A. 𝑦 =
1
2𝑥−3
» Differentiate the following functions using methods learned in
sections 2.2 & 2.3, and then by using the Chain Rule.
3
A. 𝑦 = 𝑥 + 2
We will begin by multiplying out the entire polynomial:
𝑦 = 𝑥+2 3 = 𝑥+2 𝑥+2 𝑥+2
= 𝑥 + 2 𝑥 2 + 4𝑥 + 4
= 𝑥 3 + 6𝑥 2 + 12𝑥 + 8
𝑦 ′ = 𝐷𝑥 𝑥 3 + 6𝑥 2 + 12𝑥 + 8
= 3𝑥 2 + 12𝑥 + 12
Once multiplied out the derivative is easily completed, but getting
there can be time consuming, especially for higher powers.
» Differentiate the following functions using methods learned in
sections 2.2 & 2.3, and then by using the Chain Rule.
3
B. 𝑦 = 𝑥 + 2
Now for the Chain Rule:
We begin by decomposing the two functions like in Example 1:
𝑢 =𝑥+2
𝑓(𝑢) = 𝑢3
From the Chain Rule, the derivative is:
𝑦′
𝑑
=
𝑓(𝑔 𝑥 ) = 𝑓 ′ 𝑔 𝑥 𝑔′ 𝑥 = 𝑓 ′ 𝑢 𝑢′
𝑑𝑥
So, to continue we must differentiate both of the decomposed functions.
» Differentiate the following functions using methods learned in
sections 2.2 & 2.3, and then by using the Chain Rule.
3
B. 𝑦 = 𝑥 + 2
By differentiating each we get:
𝑢=𝑥
𝒖
𝒙+2
𝟐
𝒖′ = 𝟏
𝒇′(𝒖) = 𝟑𝒖𝟐
𝑓(𝑢) = 𝑢3
So, the combined result is:
𝑦 ′ = 𝒇′ 𝒖 𝒖′ = 𝟑𝒖𝟐 𝟏 = 𝟑𝒖𝟐 = 3 𝒙 + 𝟐
2
Which when multiplied out is (which isn’t always necessary):
𝑦′ = 3 𝑥 + 2
2
= 3 𝑥 2 + 4𝑥 + 4 = 3𝑥 2 + 12𝑥 + 12
» Differentiate the following functions using methods learned in
sections 2.2 & 2.3, and then by using the Chain Rule.
B. 𝑦 = sin(2𝑥)
To differentiate this without the Chain Rule requires using a
double-angle trigonometric identity with the Product Rule:
𝑦 = sin(2𝑥) = 2 sin 𝑥 cos(𝑥)
𝑑
𝑑
′
𝑦 =2
sin(𝑥) cos 𝑥 + sin(𝑥)
cos(𝑥)
𝑑𝑥
𝑑𝑥
= 2 cos 𝑥 cos 𝑥 + sin(𝑥) −sin(𝑥)
= 2 cos2 𝑥 − sin2 (𝑥)
» Differentiate the following functions using methods learned in
sections 2.2 & 2.3, and then by using the Chain Rule.
B. 𝑦 = sin(2𝑥)
Now for the Chain Rule:
By decomposition we get:
By differentiating we get:
𝑢 = 2𝑥
𝑓(𝑢) = sin(𝑢)
𝑢′ = 2
𝑓′(𝑢) = cos(𝑢)
So, the result is:
𝑦 ′ = 𝑓 ′ 𝑢 𝑢′
= cos(𝑢) 2 = 2cos(2𝑥)
Which by another double-angle identity is equivalent to:
𝑦 ′ = 2 cos 2𝑥 = 2 cos2 𝑥 − sin2 (𝑥)
» Differentiate the following functions using methods learned in
sections 2.2 & 2.3, and then by using the Chain Rule.
C. 𝑦 =
2
3𝑥+1
We first differentiate using the Quotient Rule:
𝑑
𝑑
2 3𝑥 + 1 − 2
3𝑥 + 1
𝑑𝑥
𝑑𝑥
′
𝑦 =
3𝑥 + 1 2
=
0 3𝑥 + 1 − 2 3
6
=
−
3𝑥 + 1 2
3𝑥 + 1
2
» Differentiate the following functions using methods learned in
sections 2.2 & 2.3, and then by using the Chain Rule.
C. 𝑦 =
2
3𝑥+1
= 𝑓(𝑔 𝑥 )
Now for the Chain Rule:
By decomposition we get:
By differentiating we get:
𝑔 𝑥 = 𝑢 = 3𝑥 + 1
2
𝑓 𝑢 =
𝑢
𝑔′(𝑥) = 𝑢′ = 3
So, the result is:
𝑦′
=
𝑓′
𝑢
𝑢′
𝑓′
𝑢 =
2
= − 2
𝑢
2𝑢−1
=
−2𝑢−2
6
3 =−
3𝑥 + 1
2
2
=− 2
𝑢
» In each case, we saw how three different
problems can be solved with and without the
Chain Rule. Once you get used to the Chain
Rule, it becomes much faster and easier, and
often can be done in your head. It also is
convenient to serve as a way to check answers
when a different method is requested.
If 𝑦 = 𝑢(𝑥) 𝑛 , where 𝑢 is a differentiable
function of 𝑥 and 𝑛 is a rational number, then
𝑑𝑦
= 𝑛 𝑢(𝑥)
𝑑𝑥
𝑛−1
𝑑𝑢
𝑑𝑥
or, equivalently,
𝑑 𝑛
𝑢 = 𝑛𝑢𝑛−1 𝑢′ .
𝑑𝑥
A. Find 𝑑𝑦/𝑑𝑥 for 𝑦 = 2𝑥 2 + 5 3 .
𝑢=
2𝑥 2
𝑦 = 𝑢3
+5
𝑑𝑢
𝑢 =
= 4𝑥
𝑑𝑥
𝑑𝑦
′
𝑦 =
= 3𝑢2
𝑑𝑢
𝑑𝑦
= 3 2𝑥 2 + 5
𝑑𝑥
𝑑𝑦
𝑑𝑢
′
2
𝑑𝑦 𝑑𝑦 𝑑𝑢
=
∙
𝑑𝑥 𝑑𝑢 𝑑𝑥
4𝑥 = 12𝑥 2𝑥 2 + 5
𝑑𝑢
𝑑𝑥
2
1
𝑥 2 +1
B. If 𝑓(𝑥) =
and 𝑔(𝑥) = 𝑥, then the derivative
of 𝑓(𝑔(𝑥)) is
a)
− 𝑥
𝑥 2 +1 2
b) − 𝑥 + 1
c)
d)
e)
−2𝑥
𝑥 2 +1 2
1
𝑥+1 2
1
2 𝑥 𝑥+1
−2
This question is similar to a possible
AP multiple choice question. It is
unique in that it separates the
functions being composed.
1
𝑥 2 +1
B. If 𝑓(𝑥) =
and 𝑔(𝑥) = 𝑥, then find the
derivative of 𝑓(𝑔(𝑥)).
𝑢 =𝑥+1
We will start by saying:
ℎ 𝑥 =𝑓 𝑔 𝑥
=
1
𝑥
2
+1
=
1
.
𝑥+1
1
ℎ(𝑢) =
𝑢
However, we will not use the Chain Rule on 𝑓 and 𝑔,
but rather make a different decomposition.
1
𝑥 2 +1
B. If 𝑓(𝑥) =
and 𝑔(𝑥) = 𝑥, then find the
derivative of 𝑓(𝑔(𝑥)).
𝑓 𝑔 𝑥
=ℎ 𝑥 =
𝑢 =𝑥+1
𝑑𝑢
𝑢 =
=1
𝑑𝑥
′
1
𝑥+1
1
ℎ(𝑢) =
𝑢
𝑑ℎ
𝑑 −1
′
ℎ 𝑢 =
=
𝑢
𝑑𝑢 𝑑𝑢
= −𝑢−2 = −
1
𝑢2
1
𝑥 2 +1
B. If 𝑓(𝑥) =
and 𝑔(𝑥) = 𝑥, then find the
derivative of 𝑓(𝑔(𝑥)).
𝑓 𝑔 𝑥
=ℎ 𝑥 =
1
𝑥+1
1
𝑢 = 𝑥 + 1 ℎ(𝑢) =
𝑢
𝑑𝑢
=1
𝑑𝑥
𝑑ℎ
1
=− 2
𝑑𝑢
𝑢
𝑑ℎ 𝑑ℎ 𝑑𝑢
1
1
1
=
∙
=− 2 1 =− 2=−
𝑑𝑥 𝑑𝑢 𝑑𝑥
𝑢
𝑢
𝑥+1
We are now ready to answer the multiple choice question.
2
B. If 𝑓(𝑥) =
1
𝑥 2 +1
of 𝑓(𝑔(𝑥)) is
a)
− 𝑥
𝑥 2 +1 2
b) − 𝑥 + 1
c)
d)
e)
−2𝑥
𝑥 2 +1 2
1
𝑥+1 2
1
2 𝑥 𝑥+1
−2
and 𝑔(𝑥) = 𝑥, then the derivative
𝑑
1
𝑓(𝑔 𝑥 ) = −
𝑑𝑥
𝑥+1
Even if you solved the problem
correctly, you could easily miss
this question by not paying
attention. Notice that d) looks
like the correct answer, but it
lacks the negative sign. So, the
correct answer is not in the same
format, but by looking closely we
see the answer is b).
2
C. If
𝑑
𝑓(𝑥)
𝑑𝑥
= 𝑔(𝑥) and ℎ(𝑥) = sin(𝑥), then
𝑑
𝑓(ℎ(𝑥))
𝑑𝑥
equals
a) 𝑔 sin 𝑥
b) cos 𝑥 ∙ 𝑔(𝑥)
c) 𝑔′(𝑥)
d) cos 𝑥 ∙ 𝑔 sin 𝑥
e) sin 𝑥 ∙ 𝑔(sin 𝑥)
𝑑
𝑓 𝑥 = 𝑓′(𝑥) = 𝑔(𝑥)
𝑑𝑥
𝑑
𝑓 ℎ 𝑥
𝑑𝑥
= 𝑓 ′ ℎ 𝑥 ℎ′(𝑥)
= 𝑔 ℎ 𝑥 ℎ′ 𝑥
= 𝑔(sin 𝑥) ∙ cos 𝑥
» Using Leibniz’s Notation in the previous
example shows us how the Chain Rule can be
extended to more functions.
𝑑𝑦 𝑑𝑦 𝑑𝑢 𝑑𝑣
=
∙
∙
𝑑𝑥 𝑑𝑢 𝑑𝑣 𝑑𝑥
» The next example will demonstrate this further.
» Differentiate using the Chain Rule twice.
𝑦 = 1 + 𝑥2 − 5
𝑣 = 𝑥2 − 5
15 10
𝑢 = 1 + 𝑣 15
𝑦 = 𝑢10
𝑑𝑣
𝑑𝑢
𝑑𝑦
14
= 2𝑥
= 15𝑣
= 10𝑢9
𝑑𝑥
𝑑𝑣
𝑑𝑢
𝑑𝑦 𝑑𝑦 𝑑𝑢 𝑑𝑣
=
∙
∙
= 10𝑢9 15𝑣 14 2𝑥
𝑑𝑥 𝑑𝑢 𝑑𝑣 𝑑𝑥
= 300𝑥𝑢9 𝑣 14 = 300𝑥 1 + 𝑣 15 9 𝑣 14
= 300𝑥 1 + 𝑥 2 − 5
15 9
𝑥2 − 5
14
3
» Find all points on the graph of 𝑓(𝑥) =
𝑥2 − 4 2
for which 𝑓′(𝑥) = 0 and those for which 𝑓′(𝑥) does
not exist.
𝑓(𝑥) =
𝑓′
3
𝑥2 − 4
2 2
𝑥 = 𝑥 −4
3
2
=
1
−3
𝑥2
−4
2𝑥 =
2
3
4𝑥
3
3 𝑥2 − 4
Since the derivative is a rational function, to find where 𝑓′(𝑥) = 0, we
set the numerator equal to zero and solve. To find where 𝑓′(𝑥) does
not exist, we set the denominator equal to zero and solve.
3
» Find all points on the graph of 𝑓(𝑥) =
𝑥2 − 4 2
for which 𝑓′(𝑥) = 0 and those for which 𝑓′(𝑥) does
not exist.
𝑓′
𝑥 =
To find
4𝑥
3
3 𝑥2 − 4
𝑓′
𝑥 = 0:
4𝑥 = 0
𝑥=0
To find 𝑓 ′ 𝑥 = DNE:
3
3 𝑥2 − 4 ≠ 0
𝑥2 − 4 ≠ 0
𝑥−2 𝑥+2 ≠0
𝑥 ≠ ±2
» By studying the graphs of the original function and its
derivative, it becomes more apparent.
y
𝑓(𝑥) =
𝑓′ 𝑥 =
3
𝑥2 − 4
4𝑥
3
𝑥2 − 4
2
10
9
𝑓′(0) = 0
8
𝑥 ≠ ±2
6
You may recall that function is
NOT differentiable at a sharp
corner. This is indicative of the
vertical asymptotes at 𝑥 = ±2.
Also note the slope of the blue
graph is zero precisely where the
green crosses the x-axis, at 𝑥 = 0.
7
𝑓(𝑥)
5
4
3
2
1
–10 –9
–8
–7
–6
–5
–4
–3
–2
–1
–1
–2
𝑓′(𝑥)
–3
–4
–5
–6
–7
–8
–9
–10
1
2
3
4
5
6
7
8
9
10
x
A. Differentiate 𝑔 𝑥 =
3
.
2
5−3𝑥
𝑔 𝑥 = 3 5 − 3𝑥
𝑔′
𝑥 = −6 5 − 3𝑥
−3
−2
18
−3 =
5 − 3𝑥
Here we see it is much easier to apply the Chain Rule
and differentiate than to use the Quotient Rule. On the
next slide we will modify the function slightly, and still
use the Chain Rule.
3
B. Differentiate 𝑔 𝑥 =
𝑔 𝑥 = 3𝑥 5 − 3𝑥
𝑔′ 𝑥 = 3 5 − 3𝑥
−2
−2
3
=
5 − 3𝑥
3𝑥
.
2
5−3𝑥
Because the numerator is not a
constant, if we try to apply the
Chain Rule here, we will also have
to use the Product Rule.
+ 3𝑥 −2 5 − 3𝑥
18𝑥
+
2
5 − 3𝑥
3
−3
−3
B. Differentiate 𝑔 𝑥 =
3𝑥
.
2
5−3𝑥
Now to differentiate it with the Quotient Rule.
𝑔′
=
𝑥 =
3 5 − 3𝑥
3 5 − 3𝑥
2
− 3𝑥 2 5 − 3𝑥 −3
5 − 3𝑥 2 2
2
+ 18𝑥 5 − 3𝑥
5 − 3𝑥 4
3 5 − 3𝑥 2 18𝑥 5 − 3𝑥
=
+
4
5 − 3𝑥
5 − 3𝑥 4
3
18𝑥
=
+
2
5 − 3𝑥
5 − 3𝑥 3
Clearly for this particular
problem, the Chain Rule
combined with the Product
Rule was easier. Sometimes if
you get stuck, try a different
method.
» On the last example we just saw a problem whose derivative
was given as two different fractions:
𝑔′
3
𝑥 =
5 − 3𝑥
18𝑥
+
2
5 − 3𝑥
3
» Many times though, for instance on the multiple choice
section of the AP test, an answer may be given as one
fraction only. The following three examples illustrate
techniques for simplifying derivatives.
» Note, an alternative method would be to get a common
denominator after getting two separate fractions.
» Differentiate and simplify by factoring out the least powers.
𝑓
𝑔
1
2
2
𝑓(𝑥) = 2𝑥 9 − 𝑥 = 2𝑥 2 9 − 𝑥 2 2
𝑓′
𝑓′
𝑔
𝑥 = 4𝑥 9 −
1
2
𝑥 2
𝑓
+
2𝑥 2
Observe, when multiplying like bases,
add the exponents. We will use this
result backwards to factor.
1
1
−
−
9 − 𝑥2 2 9 − 𝑥2 1 = 9 − 𝑥2 2
𝑔′
1
2
9−
1
−
2
𝑥 2
𝑢′
𝑢
+
9
2
− 𝑥2 2
=
= 9
−2𝑥
1
− 𝑥2 2
» We will now factor out the least power, as well as the GCF.
𝑓 ′ 𝑥 = 4𝑥 9 −
= 2𝑥 9
𝟏
−
− 𝑥2 𝟐
1
𝑥2 2
+ 2𝑥 2
2 𝟗−
𝒙𝟐
1
2
9
𝟏
−
− 𝑥2 𝟐
1
+𝑥
2
GCF
9
1
2 −2
−𝑥
𝟐 𝟏
𝟗−𝒙
= 9−
1
2 2
𝑥
−2𝑥
−2𝑥
» Now to simplify.
𝑓′(𝑥) = 2𝑥 9 −
1
−
𝑥2 2
= 2𝑥 9 − 𝑥
=
2 9−
1
−
2 2
2𝑥 18 − 3𝑥 2
9−
1
𝑥2 2
𝑥2
1
+𝑥
2
18 − 2𝑥 2 − 𝑥 2
=
6𝑥 6 − 𝑥 2
9 − 𝑥2
−2𝑥
» Differentiate and simplify.
4𝑥
4𝑥
=
ℎ 𝑥 =3
𝑥2 + 1
𝑥2 + 1
𝑓′
ℎ′(𝑥) =
𝑔
4 𝑥2 + 1
𝑓
1
3
𝑓(𝑥)
1
3
𝑔(𝑥)
𝑔′
1
− 4𝑥
3
𝑥2 + 1
𝑔2
𝑥2 + 1
2
3
2
−3
𝑢
2𝑥
𝑢′
» Now to simplify the derivative.
First observe, when multiplying like bases, add the exponents.
𝑥2
ℎ′ 𝑥 =
4
𝑥2
𝑥2
+1
1
− 4𝑥 3
𝑥2
+1
+1
1
3
2
−
3
𝑥2
+1
2
3
Factored from top.
4
= 𝑥2 + 1
3
2
−3
4
= 𝑥2 + 1
3
2
−3
1
=
𝑥2
+1
+1
2
−
3
2𝑥
4÷
+1
𝑥2 + 3
𝑥2
+1
2
3
2
3
=
𝑥2
+1
+
3
3
=
𝑥2
+1
1
3
We will use this result backwards
to factor the derivative.
4
=3
3
3 𝑥 2 + 1 − 𝑥 2𝑥
𝑥2
2
−
3
4 2
= 𝑥 +1
3
4 𝑥2 + 3
3
𝑥2
+1
4
3
2
−3
3𝑥 2 + 3 − 2𝑥 2
𝑥2
+1
2
3
» Differentiate and simplify.
2
1+𝑥
𝑦=
2 + 𝑥2
1+𝑥 𝑑 1+𝑥
𝑦 =2
2 + 𝑥 2 𝑑𝑥 2 + 𝑥 2
′
1+𝑥
=2
2 + 𝑥2
1+𝑥
=2
2 + 𝑥2
1 2 + 𝑥 2 − 1 + 𝑥 2𝑥
2 + 𝑥2 2
−𝑥 2 − 2𝑥 + 2
−2 1 + 𝑥 𝑥 2 + 2𝑥 − 2
=
2 + 𝑥2 2
2 + 𝑥2 3
» Differentiate the following trigonometric
functions.
A. y = cos 5𝑥
𝑦 ′ = − sin 5𝑥
𝑑
5𝑥 = −sin(5𝑥) 5 = −5sin(5𝑥)
𝑑𝑥
B. 𝑦 = tan 4𝑥 − 1
𝑦′
=
sec 2
𝑑
4𝑥 − 1
4𝑥 − 1 = sec 2 4𝑥 − 1 4 = 4 sec 2 4𝑥 − 1
𝑑𝑥
» Parentheses change a lot with respect to
trigonometric functions. Keep that in mind
when differentiating.
A. y = sin 2𝑥 3 = sin 2𝑥 3
𝑦 ′ = cos 2𝑥 3
𝑑
2𝑥 3 = cos2𝑥 3 6𝑥 2 = 6𝑥 2 cos2𝑥 3
𝑑𝑥
B. 𝑦 = sin2
𝑦 ′ = sin 2
𝑥3
Here sin2 acts as a constant since the
variable is not inside the sine function.
𝑑 3
𝑥 = sin 2 3𝑥 2 = 3 sin 2 𝑥 2
𝑑𝑥
» Parentheses change a lot with respect to
trigonometric functions. Keep that in mind
when differentiating.
C. y = sin 2𝑥
𝑦 ′ = cos 8𝑥 3
3
= sin 8𝑥 3
𝑑
8𝑥 3 = cos8𝑥 3 24𝑥 2 = 24𝑥 2 cos8𝑥 3
𝑑𝑥
D. 𝑦 = sin2 𝑥 = sin 𝑥
2
= sin 𝑥 sin 𝑥
𝑑
𝑑
sin 𝑥 sin 𝑥 + sin 𝑥
sin 𝑥 = cos 𝑥 sin 𝑥 + sin 𝑥 cos 𝑥
𝑑𝑥
𝑑𝑥
= 2 sin 𝑥 cos 𝑥 = sin 2𝑥 The last step used a double-angle identity.
𝑦′ =
» Parentheses change a lot with respect to
trigonometric functions. Keep that in mind
when differentiating.
E. 𝑦 = sin 𝑥 = sin 𝑥
𝑦′
1
= sin 𝑥
2
1
−2
𝑑
sin 𝑥 =
𝑑𝑥
1
2
1
2 sin 𝑥
1 ∙ cos 𝑥 =
2
cos 𝑥
2 sin 𝑥
» Some trigonometric functions require repeated use of
the Chain Rule. Differentiate the following:
𝑓 𝑡 = cos3 2𝑡 = cos 2𝑡
𝑓′(𝑡) = 3 cos 2𝑡
2
= 3 cos 2𝑡
2
= 3 cos 2𝑡
2
3
𝑑
cos 2𝑡
𝑑𝑥
𝑑
− sin 2𝑡
2𝑡
𝑑𝑥
− sin 2𝑡 2
= −6cos2 2𝑡 sin 2𝑡
» Find an equation of the tangent line to the graph of
𝜋
, −1
2
𝑓 𝑥 = 3 cos 𝑥 + sin 3𝑥 at the point
. Then
determine all values of 𝑥 in the interval (0, 2𝜋) at which
the graph of 𝑓 has a horizontal tangent.
First find the derivative and substitute in 𝑥 =
𝜋
2
to find the slope at that point.
𝑓 ′ 𝑥 = −3 sin 𝑥 + 3 cos 3𝑥
𝑓
′
𝜋
𝜋
3𝜋
= −3 sin + 3 cos
2
2
2
= −3 1 + 3 0 = −3
» Find an equation of the tangent line to the graph of
𝜋
, −1
2
𝑓 𝑥 = 3 cos 𝑥 + sin 3𝑥 at the point
. Then
determine all values of 𝑥 in the interval (0, 2𝜋) at which
the graph of 𝑓 has a horizontal tangent.
𝑚=
𝑓′
𝜋
= −3
2
Now use point-slope
form to write the
equation substituting
in the point and the
slope we just found.
𝑦 − 𝑦1 = 𝑚 𝑥 − 𝑥1
𝜋
𝑦 − −1 = −3 𝑥 −
2
3𝜋
𝑦 + 1 = −3𝑥 +
2
3𝜋
𝑦 = −1 − 3𝑥 +
2
» Find an equation of the tangent line to the graph of
𝜋
, −1
2
𝑓 𝑥 = 3 cos 𝑥 + sin 3𝑥 at the point
. Then
determine all values of 𝑥 in the interval (0, 2𝜋) at which
the graph of 𝑓 has a horizontal tangent.
Horizontal tangents occur where the slope is zero, so one method to
find them is to graph the derivative and find the zeros of the function.
𝑓 ′ 𝑥 = −3 sin 𝑥 + 3 cos 3𝑥 = 0
Set the window sizes as shown to
maximize the graph. Note the x-min
and x-max are 0 & 2𝜋.
» Find an equation of the tangent line to the graph of
𝜋
, −1
2
𝑓 𝑥 = 3 cos 𝑥 + sin 3𝑥 at the point
. Then
determine all values of 𝑥 in the interval (0, 2𝜋) at which
the graph of 𝑓 has a horizontal tangent.
Horizontal tangents occur where the slope is zero, so one method to
find them is to graph the derivative and find the zeros of the function.
𝑓 ′ 𝑥 = −3 sin 𝑥 + 3 cos 3𝑥 = 0
With a little playing around with the trace
feature, it becomes obvious that the zeros
𝜋
are certain multiples of 8 .
𝜋 5𝜋 3𝜋 9𝜋 13𝜋 7𝜋
𝑥= ,
,
,
,
,
8 8 4 8 8 4
» For Example 14A-G, the following table of
functions and their derivatives is given.
Questions like these may be on the AP test.
𝑥
𝑓
𝑓′
𝑔
𝑔′
0
2
1
5
−4
1
3
2
3
−3
2
5
3
1
−2
3
10
4
0
−1
A. If 𝐴 = 𝑓 + 2𝑔, then 𝐴′(3) =
𝑥
𝑓
𝑓′
𝑔
𝑔′
0
2
1
5
−4
First we must differentiate 𝐴
to find the general function 𝐴′ .
1
3
2
3
−3
𝐴′ = 𝑓′ + 2𝑔′
2
5
3
1
−2
3
10
4
0
−1
So, 𝐴′ 3 = 𝑓 ′ 3 + 2𝑔′ 3 = 4 + 2 −1 = 2.
B. If 𝐵 = 𝑓 ∙ 𝑔, then 𝐵′(2) =
𝑥
𝑓
𝑓′
𝑔
𝑔′
0
2
1
5
−4
1
3
2
3
−3
2
5
3
1
−2
3
10
4
0
−1
First we must differentiate 𝐵 to
find the general function 𝐵′ .
𝐵′ = 𝑓 ′ 𝑔 + 𝑓𝑔′
So, 𝐵′ 2 = 𝑓 ′ 2 𝑔 2 + 𝑓 2 𝑔′ 2
= 3 1 + 5 −2 = −7.
C. If 𝐷 =
1
,
𝑔
then 𝐷′(1) =
𝑥
𝑓
𝑓′
𝑔
𝑔′
0
2
1
5
−4
1
3
2
3
−3
2
5
3
1
−2
3
10
4
0
−1
So,
𝐷′
1 =−
𝑔′ 1
𝑔 1 2
=
First we must differentiate 𝐷 to
find the general function 𝐷′ .
𝐷 = 𝑔 −1
𝐷′ = − 𝑔 −2 𝑔′
𝑔′
=− 2
𝑔
−3
− 2
3
3
9
= =
1
.
3
D. If 𝐻(𝑥) =
𝑓(𝑥), then 𝐻′(3) =
𝑥
𝑓
𝑓′
𝑔
𝑔′
0
2
1
5
−4
1
3
2
3
−3
2
5
3
1
−2
3
10
4
′
So, 𝐻 3 =
𝑓′ 3
2 𝑓(3)
First we must differentiate 𝐻 to
find the general function 𝐻′ .
1
2
𝐻(𝑥) = 𝑓(𝑥)
1
1
−
0
−1
𝐻′ 𝑥 = 𝑓 𝑥 2𝑓 ′ 𝑥
2
𝑓′(𝑥)
=
4
2
=
=
.
2 𝑓(𝑥)
2 10
10
E. If 𝐾(𝑥) =
𝑓
𝑔
(𝑥), then 𝐾′(0) =
𝑥
𝑓
𝑓′
𝑔
𝑔′
0
2
1
5
−4
1
3
2
3
−3
2
5
3
1
−2
3
10
4
0
−1
′
So, 𝐾 0 =
𝑓 ′ 0 𝑔 0 −𝑓 0 𝑔′ (0)
𝑔(0) 2
First we must differentiate 𝐾 to
find the general function 𝐾 ′ .
′ 𝑔 − 𝑓𝑔′
𝑓
𝐾′ =
𝑔2
=
1 5 −2 −4
5 2
=
13
.
25
F. If 𝑀(𝑥) = 𝑓 𝑔 𝑥 , then 𝑀′(1) =
𝑥
𝑓
𝑓′
𝑔
𝑔′
0
2
1
5
−4
1
3
2
3
−3
2
5
3
1
−2
3
10
4
0
−1
First we must differentiate 𝑀
to find the general function 𝑀′ .
𝑀′ (𝑥) = 𝑓 ′ 𝑔 𝑥 𝑔′(𝑥)
So, 𝑀′ 1 = 𝑓 ′ 𝑔 1 𝑔′ 1 = 𝑓 ′ 3 −3
= 4 −3 = −12.
G. If 𝑃(𝑥) = 𝑓 𝑥 3 , then 𝑃′(1) =
𝑥
𝑓
𝑓′
𝑔
𝑔′
0
2
1
5
−4
First we must differentiate 𝑃 to
find the general function 𝑃′ .
1
3
2
3
−3
𝑃′ (𝑥) = 𝑓 ′ 𝑥 3 3𝑥 2
2
5
3
1
−2
3
10
4
0
−1
So, 𝑃′ 1 = 𝑓 ′ 13 3 1
2
= 2 3 = 6.
= 𝑓′ 1 3