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Contents
1 Chapter 1
1
2 Derivatives
2.1 The Limit Definition of a Derivative . . . . . . . . . . . . . . . . . . .
2.1.1 Estimating the slope of a curve . . . . . . . . . . . . . . . . .
2.1.2 The Limit Definition of a the Derivative . . . . . . . . . . . .
2.1.3 Finding the Derivative of a Function . . . . . . . . . . . . . .
2.2 The Power Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.1 Introduction to the Power Rule . . . . . . . . . . . . . . . . .
2.2.2 The Slope of a Tangent Line . . . . . . . . . . . . . . . . . . .
2.2.3 Derivative of the Trigonometric Functions . . . . . . . . . . .
2.2.4 The Derivative of the Exponential Function . . . . . . . . . .
2.2.5 Applications of the Derivative . . . . . . . . . . . . . . . . . .
2.3 The Product and Quotient Rule . . . . . . . . . . . . . . . . . . . . .
2.3.1 The Product Rule . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.2 Tangent Lines . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.3 The Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . .
2.4 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.1 General Chain Rule . . . . . . . . . . . . . . . . . . . . . . . .
2.4.2 The Chain Rule and Trigonometric Functions . . . . . . . . .
2.4.3 The Chain Rule for Exponential functions . . . . . . . . . . .
2.4.4 The Derivative of the Natural Logarithm Function . . . . . . .
2.5 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.1 Examples of Implicit Derivatives . . . . . . . . . . . . . . . .
2.5.2 Implicit Exponential and Trigonometric Functions . . . . . . .
2.5.3 Tangent Lines . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.4 Finding the Derivatives of the inverse Trigonometric Functions
Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . .
2.6 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.1 The Inverse Trigonometric Functions . . . . . . . . . . . . . .
2.6.2 Finding the Derivatives of the Inverse Trigonometric Functions
2.7 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.1 Applications of Related Rates . . . . . . . . . . . . . . . . . .
1
Chapter 1
See Math 151 Notes
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2
2.1
2.1.1
Derivatives
The Limit Definition of a Derivative
Estimating the slope of a curve
The main goal of this section is to find ways to measure rates of change. In Algebra, the
rate is measured by the slope of linear equation. In Calculus, we are interest in measure the
slope of a curve. In this chapter, we are going to investigate how to measure the slope of the
tangent line to the curve.
The problem we have with the slope of a tangent line to a curve is we only know one set of
coordinates where the line and curve intersect. The slope formula requires two points on the
1
line. Recall that the formula for slope is m = xy22 −y
What can we can do is find the slope of
−x1
a secant line that intersects the curve in two places.
2
2.1.2
The Limit Definition of a the Derivative
If we place the two points on the secant line as close together as we can, we can estimate
the slope of the secant line which will be close in value to the slope of the tangent at that
point. Using this idea along with the slope formula, we can derive a formula to estimate the
slope of the tangent line.
In order to get the points (x, f (x)) and (x + h, f (x + h)) as close getter as possible, we can
take the limit as h approaches zero of the slope of the line passing through the two points.
Recall that h is the distance between the two points.
m=
y2 −y1
f (x + h) − f (x)
f (x + h) − f (x)
= lim
= lim
x2 −x1 h→0 x + h − x
h→0
h
The formula we just found is known as the limit definition of a derivative. We will use this
formula to find the derivative of a function.
The Limit Definition of a Derivative
f (x + h) − f (x)
h→0
h
f ′ (x) = lim
2.1.3
Finding the Derivative of a Function
Now, let’s use the limit definition of a derivative to find the derivative of a function.
Example 1
Find the derivative of the function using the limit definition of a derivative. f (x) = 3x − 2
f (x + h) − f (x)
h
3(x + h) − 2 − (3x − 2)
= lim
h→0
h
3x + 3h − 2 − 3x + 2
= lim
h→0
h
f ′ (x) = lim
h→0
3
3h
h→0 h
= lim 3 = 3
= lim
h→0
Example 2
Find the derivative of the function using the limit definition of a derivative. f (x) = 5x + 3
f (x + h) − f (x)
h→0
h
5(x + h) + 3 − (5x + 3)
= lim
h→0
h
5x + 5h + 5 − 5x − 3
= lim
h→0
h
5h
= lim
= lim 5 = 5
h→0 h
h→0
f ′ (x) = lim
Example 3
Find the derivative of the function using the limit definition of a derivative. f (x) = x2 − 6x
f (x + h) − f (x)
h→0
h
(x + h)2 − 6(x + h) − (x2 − 6x)
= lim
h→0
h
(x + h)(x + h) − 6x − 6h − x2 + 6x
= lim
h→0
h
x2 + 2xh + h2 − 6x − 6h − x2 + 6x
= lim
h→0
h
2xh + h2 − 6h
= lim
h→0
h
h(2x + h − 6)
= lim
h→0
h
= lim (2x + h − 6)
h→0
= 2x + 0 − 6
= 2x − 6
f ′ (x) = lim
Note That: (x + h)(x + h) = x · x + x · h + x · h + h · h = x2 + 2xh + h2
Example 4
Find the derivative of the function using the limit definition of a derivative. f (x) = x2 + 4
f (x + h) − f (x)
h→0
h
(x + h)2 + 4 − (x2 + 4)
= lim
h→0
h
(x + h)(x + h) + 4 − x2 − 4
= lim
h→0
h
x2 + 2xh + h2 + 4 − x2 − 4
= lim
h→0
h
f ′ (x) = lim
4
2xh + h2
h→0
h
h(2x + h)
= lim
h→0
h
= lim (2x + h)
h→0
= 2x + 0 = 2x
= lim
Example 5
Find the derivative of the function using the limit definition of a derivative. f (x) = x3 + 2
f (x + h) − f (x)
h→0
h
(x + h)3 + 2 − (x3 + 2)
= lim
h→0
h
(x + h)(x + h)(x + h) + 2 − x3 − 2
= lim
h→0
h
(x + h)(x2 + 2xh + h2 ) + 2 − x3 − 2
= lim
h→0
h
x3 + 3x2 h + 3xh2 + h3 − x3
= lim
h→0
h
3x2 h + 3xh2 + h3
= lim
h→0
h
h(3x2 + 3xh + h2 )
= lim
h→0
h
= lim (3x2 + 3xh + h2 )
f ′ (x) = lim
h→0
= 3x2 + 3x(0) + 02
= 3x2
Example 6
Find the derivative of the function using the limit definition of a derivative. f (x) = x3 − 3x
f (x + h) − f (x)
h→0
h
(x + h)3 − 3(x + h) − (x3 − 3x)
= lim
h→0
h
(x + h)(x + h)(x + h) − 3x − 3h − x3 + 3x
= lim
h→0
h
(x + h)(x2 + 2xh + h2 ) − 3x − 3h − x3 + 3x
= lim
h→0
h
x3 + 3x2 h + 3xh2 + h3 − 3h − x3
= lim
h→0
h
3x2 h + 3xh2 + h3 − 3h
= lim
h→0
h
h(3x2 + 3xh + h2 − 3)
= lim
h→0
h
= lim (3x2 + 3xh + h2 − 3)
f ′ (x) = lim
h→0
5
= 3x2 + 3x(0) + 02 − 3
= 3x2 − 3
Example 7
Find the derivative of the function using the limit definition of a derivative. f (x) =
√
x+4
f (x + h) − f (x)
f ′ (x) = lim
h √
√h→0
(x + h) + 4 − x + 4
= lim
h→0 √
h
√
√
√
(x + h) + 4 − x + 4
(x + h) + 4 + x + 4
= lim
·√
√
h→0
h
(x
+
h)
+
4
+
x+4
√
√
√
√
√
√
2
( (x + h) + 4) + ( x + h + 4)( x + 4) − ( x + h + 4)( x + 4) − ( x + 4)2
√
= lim
√
h→0
h( (x + h) + 4 + x + 4)
x + h + 4 − (x + 4)
= lim √
√
h→0 h(
(x + h) + 4 + x + 4)
x+h+4−x−4
= lim √
√
h→0 h(
(x + h) + 4 + x + 4)
h
= lim √
√
h→0 h(
(x + h) + 4 + x + 4)
1
= lim √
√
h→0
(x + h) + 4 + x + 4
1
√
=√
x+0−4+ x+4
1
√
=
2· x+4
Example 8
1
Find the derivative of the function using the limit definition of a derivative. f (x) = x+2
f (x + h) − f (x)
h→0
h
1
1
−
(x+h)+2
x+2
f ′ (x) = lim
= lim
h→0
= lim
h
1
(x+h)+2
·
x+2
x+2
h→0
= lim
x+2
(x+h+2)(x+2)
h→0
= lim
h→0
= lim
h→0
= lim
h→0
x+2−(x+h+2)
(x+h+2)(x+2)
−
1
x+2
·
(x+h)+2
(x+h)+2
h
x+h+2
− (x+h+2)(x+2)
h
h
x+2−x−h−2
(x+h+2)(x+2)
h
−h
(x+h+2)(x+2)
h
6
−1
h→0 (x + h + 2)(x + 2)
= lim
=
=
−1
(x+0+2)(x+2)
−1
(x+2)2
Example 9
Find the slope of a tangent line to the function f (x) = x2 − 3 at the point (1, −2)
Solution:
First, find the derivative of the function. Recall that the derivative measures the slope of
the tangent.
f (x + h) − f (x)
h
(x + h)2 − 3 − (x2 − 3)
= lim
h→0
h
(x + h)(x + h) − 3 − x2 + 3
= lim
h→0
h
x2 + 2xh + h2 − 3 − x2 + 3
= lim
h→0
h
2xh + h2
= lim
h→0
h
h(2x + h)
= lim
h→0
h
= lim (2x + h)
h→0
= 2x + 0 = 2x
f ′ (x) = lim
h→0
Now, find the slope of the tangent line by substituting in x = 1 into the derivative.
m = f ′ (1) = 2(1) = 2
7
8
2.2
2.2.1
The Power Rule
Introduction to the Power Rule
The next derivative technique we will discuss is the power rule. Before we learn the general
form of the power rule, let’s review some result’s from using the limit definition of a derivative
in the previous section.
f (x)
Example
Example
Example
Example
1
3
5
6
f (x) = 5x + 3
f (x) = x2 − 6x
f (x) = x3 + 2
f (x) = x3 − 3x
f ′ (x)
f ′ (x) = 5
f ′ (x) = 2x − 6
f ′ (x) = 3x2
f ′ (x) = 3x2 − 3
If we study the patterns from the basic function to its derivative, we see that we can find the
derivative by multiplying the exponent by the leading coefficient of the term and subtracting
one from the exponent. The power rule also can be expressed as a simple formula.
The Power Rule
d
(cxn )
dx
= n · cxn−1
Next, we will formally prove the power rule by using the binomial expansion of a polynomial.
Proof of the Power Rule
Define the function f (x) = xx and use the limit definition of a derivative to find the derivative
of f (x).
f (x + h) − f (x)
h→0
h
(x + h)n − xn
′
f (x) = lim
h→0
h
n
n−1
[x
+
nx
h + n(n−1)
xn−2 h2 + ..... + nxhn−1 + hn ] − xn
2
′
f (x) = lim
h→0
h
n(n−1) n−2 2
n−1
nx
h
+
x
h
+
.....
+ nxhn−1 + hn
2
f ′ (x) = lim
h→0
h
n(n−1) n−2
n−1
h(nx
+ 2 x h + ..... + nxhn−2 + hn−1 )
′
f (x) = lim
h→0
h
n(n − 1) n−2
n−1
′
f (x) = lim [nx
+
x h + ..... + nxhn−2 + hn−1 ]
h→0
2
= nxn−1 + 0 + 0 + ....... + 0
= nxn−1
f ′ (x) = lim
Now, let’s use the power to find the derivative of the function.
Example 1
Find the derivative of the function using the power rule.
f (x) = 2x3
9
f ′ (x) = 2x3 = 3 · 2x3−2 = 6x2
Example 2
Find the derivative of the function using the power rule.
f (x) = x4 + 4x3 − 5x2 − 8x + 5
f ′ (x) = 4x4−1 + 3 · 4x3−1 − 2 · 5x2−1 − 8x1−1 + 0
f ′ (x) = 4x3 + 12x2 − 10x − 8x0
f ′ (x) = 4x3 + 12x2 − 10x − 8
Example 3
Find the derivative of the function using the power rule.
f (x) =
1
x2
Recall; f (x) = x12 = x−2
f ′ (x) = −2x−2−1 = −2x−3 = − x23
Example 4
Find the derivative of the function using the power rule.
f (x) =
√
x
Recall: f (x) =
√
1
x = x2
f ′ (x) = 12 x 2 −1 = 12 x− 2 =
1
2.2.2
1
1
1
2x 2
=
1
√
2 x
The Slope of a Tangent Line
Example 5
Find the slope of a tangent line to the function f (x) = x2 − 3 at the point (1, −2).
Solution:
Recall that the slope of the tangent line to a function is measured by the derivative of the
function.
First find the derivative of the function. f ′ (x) = 2x2−1 − 0 = 2x
Next, substitute the value of x into the derivative. In this case the value of x in the point
(1, 2) is x = 1
10
m = f ′ (1) = 2 · 1 = 2
If we look at the graph of the function f (x) = x2 −3, we can see that the slope of the tangent
at (1, 2) is m = 2.
Example 6
Find the equation of a tangent line to the function f (x) = x3 + 2 at the point (1, 3).
Solution:
Recall that the slope of the tangent line to a function is measured by the derivative of the
function. Therefore, the first step will be to find the derivative of the function. f ′ (x) =
3x3−1 + 0 = 3x2
Next, substitute the value of x into the derivative. In this case the value of x in the point
(1, 3) is x = 1
m = f ′ (1) = 3(1)2 = 3 · 1 = 3
Next, use the point-slope formula to find the equation of the line.
y − y1 = m(x − x1 )
y − 3 = 3(x − 1)
y − 3 = 3x − 3
y − 3 + 3 = 3x − 3 + 3
y = 3x
If we look at the graph of the function f (x) = x3 +2, we can see that the slope of the tangent
at (1, 3) is m = 3.
11
2.2.3
Derivative of the Trigonometric Functions
Before we learn the derivatives of the trigonometric function we must first derive their
derivatives. We will start by deriving the derivative for the sine function. We will also
need to know two important limits before we derive the derivatives of the sine function and
the cosine function. If we look at the graph of f (h) = sin(h)
, we can determine its limit.
h
sin(h)
=1
h→0
h
Fact 1: lim
Likewise we can use the following graph of f (h) =
12
cos(h)−1
h
to show that that Fact 2: is
cos(h) − 1
=0
h→0
h
lim
We will use these two facts to derive the derivative of f (x) = sin(x)
Proof: Let f (x) = sin(x) and use the limit definition of a derivative to find the derivative
of f (x) = sin(x)
f (x + h) − f (x)
h→0
h
sin(x
+
h) − sin(x)
f ′ (x) = lim
h→0
h
sin(x)cos(h) + sin(h)cos(x) − sin(x)
′
f (x) = lim
h→0
h
sin(x)cos(h) − sin(x)
sin(h)cos(x)
′
f (x) = lim
+ lim
h→0
h→0
h
h
sin(x)(cos(h)
−
1)
sin(h)cos(x)
f ′ (x) = lim
+ lim
h→0
h→0
h
h
(cos(h)
−
1)
sin(h)
f ′ (x) = sin(x) · lim
+ cos(x) · lim
h→0
h→0
h
h
f ′ (x) = lim
Now use Fact 1: is lim
h→0
the limit.
cos(h) − 1
sin(h)
= 1 and Fact 2: is lim
= 0 from above to find
h→0
h
h
f ′ (x) = sin(x) · 0 + cos(x) · 1
f ′ (x) = 0 + cos(x)
f ′ (x) = cos(x)
we will do a similar process to find the derivative of the cosine function.
Let f (x) = cos(x) and use the limit definition of a derivative to find the derivative of
f (x) = cos(x)
13
f (x + h) − f (x)
h→0
h
cos(x
+
h) − cos(x)
f ′ (x) = lim
h→0
h
cos(x)cos(h) − sin(x)sin(h) − cos(x)
′
f (x) = lim
h→0
h
cos(x)cos(h)
−
cos(x)
sin(x)sin(h)
f ′ (x) = lim
− lim
h→0
h→0
h
h
sin(x)sin(x)
cos(x)(cos(h)
−
1)
− lim
f ′ (x) = lim
h→0
h→0
h
h
(cos(h) − 1)
sin(h)
′
f (x) = cos(x) · lim
− sin(x) · lim
h→0
h→0
h
h
f ′ (x) = lim
Again, use the Facts 1 and 2 from the previous proof to find the limit.
f ′ (x) = cos(x) · 0 − sin(x) · 1
f ′ (x) = 0 − sin(x)
f ′ (x) = −sin(x)
The proofs of tangent and cotangent will be shown later in the chapter. For now, here are
the derivatives of the six trigonometric functions.
d
(sin(x))
dx
= cos(x)
d
(cos(x))
dx
= −sin(x)
d
(tan(x))
dx
= sec2 (x)
d
(cot(x))
dx
= −csc2 (x)
d
(sec(x))
dx
= sec(x)tan(x)
d
(csc(x))
dx
= −csc(x)cot(x)
Example 7
Find the derivative of the function. f (x) = sin(x) + 4
Solution:
f ′ (x) = cos(x) + 0 = cos(x)
Example 8
Find the derivative of the function. f (x) = cos(x) + tan(x)
Solution:
f ′ (x) = −sin(x) + sec2 (x)
14
Example 9
Find the derivative of the function. f (x) = 3cos(x) + 4x2
Solution:
f ′ (x) = 3(−sin(x)) + 2 · 4x2−1 = −3cos(x) + 8x
2.2.4
The Derivative of the Exponential Function
The exponential function ex was introduced in section 1.4 along with other properties of
exponents. In the next example, we will derive the derivative of the exponential function.
Let f (x) = ex and find the derivative using the limit definition of a derivative.
f (x + h) − f (x)
h
ex+h − ex
′
f (x) = lim
h→0
h
ex · eh − ex
′
f (x) = lim
h→0
h
x h
e
(e
− 1)
f ′ (x) = lim
h→0
h
h
e
−1
f ′ (x) = ex · lim
h→0
h
f ′ (x) = ex · 1
f ′ (x) = ex
f ′ (x) = lim
h→0
eh − 1
= 1 was found by taking the limit from its graph which is shown below:
h→0
h
Note that lim
The derivative of the exponential function
15
d
(ex )
dx
= ex
Example 10
Find the derivative of y = 3ex
Solution:
y = 3ex
y ′ = 3 · ex
y ′ = 3ex
Example 11
Find the derivative of f (x) = 2x3 + sin(x) + ex
Solution:
f (x) = 2x3 + sin(x) + ex
f ′ (x) = 3 · 2x3−1 + cos(x) + ex
f ′ (x) = 6x2 + cos(x) + ex
2.2.5
Applications of the Derivative
In physics, we can use the derivative to find the velocity of an object and the second derivative
(the derivative of the first derivative) to find the acceleration of the object. Given the
displacement function s(t), the velocity and acceleration of an object is defined as follows.
v(t) = s′ (t)
a(t) = s′′ (t)
Now let’s try an example.
Example 12
The equation of motion of a particle is s(t) = t3 − 4t2 where s is in meters and t is seconds.
(A) Find the velocity and acceleration as a function of t.
(B) Find the velocity after 4 seconds.
(C) Find the acceleration after 4 seconds.
Solution:
Part A
s(t) = t3 − 4t2
v(t) = s′ (t) = 3 · t3−1 − 2 · 4t2−1 = 3t2 − 8t
a(t) = s′′ (t) = 6t − 8
16
Part B
At t = 4 ⇒ v(4) = 3(4)2 − 8(4) = 3(16) − 32 = 48 − 32 = 16 ms
Part C
At t = 4 ⇒ a(4) = 6(4) − 8 = 24 − 8 = 16 sm2
17
2.3
2.3.1
The Product and Quotient Rule
The Product Rule
The next derivative rule we will learn is called Product Rule.
The Product Rule
d
(F (x)S(x))
dx
= S ′ (x) · F (x) + F ′ (x) · S(x)
Example 1
Find the derivative of the following function.
f (x) = (x3 + 4x2 + 6)(4x3 − 5x2 )
Solution:
f (x) = (x3 + 4x2 + 6)(4x3 − 5x2 )
f ′ (x) = (4x3 − 5x2 )′ (x3 + 4x2 + 6) + (x3 + 4x2 + 6)′ (4x3 − 5x2 )
f ′ (x) = (12x2 − 10x)(x3 + 4x2 + 6) + (3x2 + 8x)(4x3 − 5x2 )
f ′ (x) = 12x5 + 48x4 + 72x2 − 10x3 − 40x2 − 60x + 12x5 − 15x4 + 32x4 − 40x3
f ′ (x) = 24x5 + 65x4 − 50x3 + 32x2
Example 2
Find the derivative of the following function.
f (x) = x5 ex
Solution:
f (x) = x5 ex
f ′ (x) = (ex )′ · x5 + (x5 )′ · ex
f ′ (x) = x5 ex + 5x4 ex
Example 3
Find the derivative of the following function.
f (x) = ex sin(x)
Solution:
f (x) = ex sin(x)
f ′ (x) = (sinx)′ · ex + (ex )′ · sin(x)
f ′ (x) = ex cos(x) + ex sin(x)
18
2.3.2
Tangent Lines
Example 4
Find the equation of a tangent line of the function f (x) = 2xex at the (0, 0)
Solution:
Find the slope of the tangent first by taking the derivative of the function and substituting
the value of x into the derivative.
f (x) = 2xex
f ′ (x) = (ex )′ · 2x + (2x)′ · ex
f ′ (x) = 2xex + 2ex
m = f ′ (0) = 2(0)e0 + 2e0 = 0 · 1 + 2 · 1 = 0 + 2 = 2
Now, use the point-slope formula to get the equation of the tangent line.
y − y1 = m(x − x1 )
y − 0 = 2(x − 0)
y = 2x
2.3.3
The Quotient Rule
The next derivative rule we will learn is called Quotient Rule.
The Quotient Rule
d T (x)
dx ( B(x) )
=
B(x)T ′ (x)−T (x)B ′ (x)
[B(x)]2
Example 5
Find the derivative of the function y =
x2
x2 +6x
Solution;
2
y = x2x+6x
(x2 +6x)(x2 )′ −(x2 )(x2 +6x)′
(x2 +6x)2
(x2 +6x)(2x)−(x2 )(2x+6)
(x2 +6x)2
2x3 +12x2 −2x3 −6x2
(x2 +6x)2
2
6x
(x2 +6x)2
y′ =
=
=
=
19
Example 6
Find the derivative of the function y =
ex
cos(x)
Solution;
x
e
y = cos(x)
(cos(x))(ex )′ −(ex )(cos(x))′
y′ =
(cos(x))2
=
=
(cos(x))(ex )−(ex )(−sin(x))
cos2 x
ex cos(x)+ex −sin(x)
cos2 x
Example 7
Use the quotient rule to show that
d
(tan(x))
dx
= sec2 (x)
Solution;
Let f (x) = tan(x)
sin(x)
f (x) = cos(x)
(cos(x))(sinx)′ −(sinx)(cos(x))′
f ′ (x) =
(cos(x))2
=
=
=
(cos(x))(cos(x))−(sin(x)(−sin(x))
cos2 x
cos2 (x)+sin2 (x)
cos2 x
1
cos2 x
= sec2 x
Example 8
Use the quotient rule to show that
d
(cot(x))
dx
= −csc2 (x)
Solution;
Let f (x) = cot(x)
cos(x)
f (x) = sin(x)
(sin(x))(cosx)′ −(cosx)(sin(x))′
f ′ (x) =
(sin(x))2
=
=
=
=
(sin(x))(−sin(x))−(cos(x))(cos(x))
sin2 x
−sin2 (x)−cos2 (x)
sin2 x
−1(sin2 (x)+cos∗2(x))
sin2 x
−1
sin2 (x)
= −csc2 (x)
20
Example 9
2
Find the slope of a tangent line to the function f (x) = x2x+3 at the point (1, 14 )
Solution;
First, find the derivative of the function.
(x2 +3)(x2 )′ −(x2 )(x2 +3)′
(x2 +3)2
2
2
(x +3)(2x)−(x )(2x)
(x2 +3)2
2x3 −2x3
(x2 +3)2
6x
(x2 +3)2
f ′ (x) =
=
=
=
Now, find the slope of the lime by substituting x = 1 into the derivative.
f ′ (1) =
6(1)
(12 +3)2
=
6
42
=
6
16
=
3
8
21
2.4
The Chain Rule
2.4.1
General Chain Rule
The chain rule is a special derivative technique used to breakdown a composition of functions.
In this case, we can find the derivative of a function that is in the form of a composition
of functions. The basic definition for the chain rule is represented by the following general
formula.
dy
dx
=
dy
du
·
du
dx
Now, let’s look at an example that it is in the form of a composition of two functions. We
will use the chain rule to find the derivative.
Example 1
Find the derivative of the function. y = (x2 + 3x)6
Solution:
y = (x2 + 3x)6
Let y = u6 where u = x2 + 3x ⇒
du
dx
= 2x + 3
Then, dx = du · du
dx
dy
6−1
· (2x + 3)
dx = 6u
dy
dy
dy
5
dx = 6u · (2x + 3)
dy
2
5
dx = (x + 3x) · (2x + 3)
dy
2
5
dx = (12x + 18)(x + 3x)
Example 2
Find the derivative of the function. y =
√
x2 + 6x + 4
Solution:
√
y = x2√
+ 6x + 4
√
1
Let y = x2 + 6x + 4 = u = u 2 where u = x2 + 6x + 4 ⇒
Then, dx = du · du
dx
dy
1 12 −1
=
u
·
(2x
+
6)
2
dx
dy
dy
dy
1 − 21
· (2x + 6)
dx = 2 u
dy
− 12
2
· (2x + 6)
dx = (x + 6x + 4)
dy
2x+6
dx = 2(x2 +6x+4) 12
dy
√ 2x+6
dx = 2 x2 +6x+4
22
du
dx
= 2x + 6
dy
dx
dy
dx
=
=
2.4.2
2(x+3)
√
2 x2 +6x+4
√ x+3
x2 +6x+4
The Chain Rule and Trigonometric Functions
Example 3
Find the derivative of the function. f (x) = sin(2x + 5)
Solution:
Let f (x) = sin(2x + 5) = sin(u) where u = 2x + 5 ⇒ du = 2
f ′ (x) = cos(u) · du
f ′ (x) = 2cos(2x + 3)
2.4.3
The Chain Rule for Exponential functions
Let y = eu , then then chain rule gives the following function.
dy
dx
=
dy
du
·
du
dx
Definition:
= eu · du
d
(eu )
dx
= eu · du
Example 4
Find the derivative of the function. y = ex
2 +5x
Solution:
Let y = eu where u = x2 + 5x ⇒
dy
dx
= 2x + 5
y ′ = eu · du = (2x + 5)ex
2 +5x
Example 5
Find the derivative of the function. f (x) = e5x sin(x)
Solution:
f (x) = e5x sin(x)
f ′ (x) = (sin(x))(e5x )′ + (e5x )(sin(x))′
f ′ (x) = (sin(x))(5e5x ) + (e5x )(cos(x))
f ′ (x) = 5e5x sin(x) + e5x cos(x)
23
Example 6
Find the derivative of the function. f (x) = x4 cos(3x2 )
Solution:
f (x) = x4 cos(3x2 )
f ′ (x) = (cos(3x2 ))(x4 )′ + (x4 )(cos(3x2 ))′
f ′ (x) = (cos(3x2 ))(4x3 ) + (x4 )(−sin(3x2 )(6x))
f ′ (x) = 4x3 cos(3x2 ) − 6x5 sin(3x2 )
2.4.4
The Derivative of the Natural Logarithm Function
The derivative of the natural logarithm function.
Let y = ln(u)
Definition:
d
(ln(x))
dx
=
1
x
Example 7
Find the derivative of the following function. f (x) = x3 + ln(x) + 3
Solution:
f (x) = x3 + ln(x) + 3
f ′ (x) = 3x3−1 + x1
f ′ (x) = 3x2 + x1
The Chain Rule for ln(u)
dy
dx
=
dy
du
·
du
dx
=
1
u
· du =
du
u
Example 8
Find the derivative of the following function. f (x) = ln(5x2 + 10x)
Solution:
f (x) = ln(5x2 + 10x)
f (x) = ln(u)
where u = 5x2 + 10x ⇒ du = 10x + 10
f ′ (x) = du
u
5(2x+2)
2x+2
10x+10
f ′ (x) = 5x
2 +10x = 5(x2 +2x) = x2 +2x
24
Example 9
Find the derivative of the following function. f (x) = e4x ln(2x)
Solution:
f (x) = e4x ln(2x)
f ′ (x) = (ln(2x))′ · e5x + (e5x )′ · ln(2x)
2
f ′ (x) = ( 2x
) · e5x + (5e5x ) · ln(2x)
5x
f ′ (x) = ex + 5e5x · ln(2x)
Example 10
Find the slope of a tangent line to the function f (x) =
e3x
2x+1
at the point (0, 1)
Solution:
First, find the derivative of the function.
e3x
f (x) = 2x+1
f ′ (x) =
f ′ (x) =
f ′ (x) =
(2x+1)(e3x )′ −(e3x )(2x+1)′
(2x+1)2
(2x+1)(3e3x )−(e3x )(2)
(2x+1)2
(6x+3)e3x −2e3x
(2x+1)2
Next, substitute the value x = 0 into the derivative to find the slope of the
tangent line.
m = f ′ (0) =
(6(0)+3)e3·0 −2e3·0
(2·0+1)2
=
(0+3)e3·0 −2e3·0
(0+1)2
25
=
3e0 −2e0
(1)2
=
3−2
1
=
1
1 =1
2.5
Implicit Differentiation
The functions we have used so far have been described in terms of one variable (Explicit
Functions).
y = x2 + 5
y = cos(x)
Recall that functions can also be described implicitly, in other words, described in terms of
two or more variables. Here are some functions that are implicit functions.
x2 + y 2 = 5
x3 + 2x2 + 5xy = 9
2.5.1
Examples of Implicit Derivatives
We use implicit differentiation to find the derivative of a function that is defined for two
variables. Here is an example of how to find the derivative using implicit differentiation.
Example 1
Find the derivative implicitly
x2 + 3y 2 = 3
First, find the derivative with respect to x. In the case with terms that contain y, we need
to apply the chain rule by multiplying by the term y’.
d
(x2 +
dx
2−1
d
3y 2 ) = dx
(3)
2−1
2·x
+3·y
· y′ = 0
′
2x + 6yy = 0
Now, solve the above equation for y’. This will result in the implicit derivative.
2x − 2x + 6yy ′ = 0 − 2x
6yy ′ = −2x
x
= − 3y
y ′ = −2x
6y
Example 2
Find the derivative implicitly
x2 + 3xy + y 2 = 5
First, find the derivative with respect to x. In the case with terms that contain y, we need
to apply the chain rule by multiplying by the term y’.
d
(x2
dx
+ 3xy + y 2 ) =
d
(5)
dx
26
2x + y ′ (3x) + (3x)′ y + 2yy ′ = 0
2x + 3xy ′ + 3y + 2yy ′ = 0
Now, solve the above equation for y’. This will result in the implicit derivative.
3xy ′ + 2yy ′ = −3y − 2x
(3x + 2y)y ′ = −(2x + 3y)
y ′ = −(2x+3y)
3x+2y
y ′ = − 2x+3y
3x+2y
Example 3
Find the derivative implicitly
x2 y + xy 2 = 9x2
First, find the derivative with respect to x. In the case with terms that contain y, we need
to apply the chain rule by multiplying by the term y’.
d
(x2 y
dx
′ 2
d
+ xy 2 ) = dx
(9x2 )
y (x ) + 2xy + (2yy ′ )x + (1)y 2 = 18x
x2 y ′ + 2xy + 2xyy ′ + y 2 = 18x
Now, solve the above equation for y’. This will result in the implicit derivative.
x2 y ′ + 2xyy ′ = 18x − 2xy − y 2
(x2 + 2xy)y ′ = 18x − 2xy − y 2
2
y ′ = 18x−2xy−y
x2 +2xy
2.5.2
Implicit Exponential and Trigonometric Functions
Example 4
Find the derivative implicitly
x2 y 2 + x2 cos(x) = 6
First, find the derivative with respect to x. In the case with terms that contain y, we need
to apply the chain rule by multiplying by the term y’.
d
(x2 y 2
dx
′
d
+ x2 cos(x)) = dx
(6)
2
′
2yy + 2xy − sin(y)y · x2 + 2x · cos(y) = 0
2yy ′ + 2xy 2 − x2 sin(y)y ′ + 2xcos(y) = 0
2yy ′ − x2 sin(y)y ′ = −2xy 2 − 2xcos(y)
(2y − x2 sin(y))y ′ = −2xy 2 − 2xcos(y)
y′ =
−2xy 2 −2xcos(y)
2y−x2 sin(y)
27
Example 5
Find the derivative implicitly
y 2 ex + 5x = 6y
First, find the derivative with respect to x. In the case with terms that contain y, we need
to apply the chain rule by multiplying by the term y’.
d
(y 2 ex +
dx
x ′
2
5x) = dd 6y
(e ) · y + (y 2 )′ · ex + 5 = 6y ′
ex y 2 + 2yex y ′ + 5 = 6y ′
ex y 2 + 5 = 6y ′ − 2yex y ′
ex y 2 + 5 = (6 − 2yex )y ′
ex y 2 +5
y ′ = 6−2yex
2.5.3
Tangent Lines
Example 6
Use implicit differentiation to find the slope of a tangent line to the curve given by 2x2 +
3xy + y 2 = 6 at the point (1, 1).
First, find the derivative with respect to x. In the case with terms that contain y, we need
to apply the chain rule by multiplying by the term y’.
d
(2x2
dx
d
+ 3xy + y 2 ) = dx
(6)
4x + y · 3x + 3y + 2yy ′ ) = 0
4x + 3xy ′ + 3y + 2yy ′ = 0
3xy ′ + 2yy ′ = −4x − 3y
(3x + 2y)y ′ = −(4x + 3y)
4x+3y
y ′ = − 3x+2y
′
Now find the slope of the tangent line by substituting x = 1 and y = 1
4(1)+3(1)
m = − 3(1)+2(1)
= − 4+3
= − 57
3+2
28
Example 7
Use implicit differentiation to find the slope of a tangent line to the curve given by x2 − xy +
y 2 = 1 at the point (1, 1).
First, find the derivative with respect to x.
d
− xy + y 2 ) = dx
(1)
′
2x − y · x − (1) · y + 2yy ′ ) = 0
2x − y ′ x + y + 2yy ′ = 0
2yy ′ − xy ′ = −2x − y
(2y − x)y ′ = −(2x − y)
2x−y
y ′ = − 2y−x
d
(x2
dx
Now find the slope of the tangent line by substituting x = 1 and y = 1
2(1)−1
m = − 2(1)−1
= − 2−1
= − 11 = −1
2+1
Example 8
Find the derivative using implicit differentiation.
2sin(x)sin(y) = 1
Solution:
d
(2sin(x)cos(y))
dx
′
d
= dx
(1)
(2cos(y)) sin(x) + (sin(x))′ · 2cos(y) = 0
−2sin(y)sin(x)y ′ + 2cos(x)cos(y) = 0
−2sin(x)sin(y)y ′ = −2cos(x)cos(y)
−2cos(x)cos(y)
y ′ = −2sin(x)cos(x) = cot(x)cot(y)
Example 9
Use implicit differentiation to find the slope of a tangent line to the curve given by x3 −y 3 = 0
at the point (1, 1).
First, find the derivative with respect to x. In the case with terms that contain y, we need
to apply the chain rule by multiplying by the term y’.
d
− y 3 ) = dx
(0)
2 ′
3x − 3y y ) = 0
−3y 2 y ′ = −3x2
2
x2
y ′ = −3x
3y 2 = y 2
d
(x3
dx
2
Now find the slope of the tangent line by substituting x = 1 and y = 1
2
m = − 112 = − 11 = 1
29
Example 10
Find the derivative using implicit differentiation.
sin(x) = 4cos(3y) = 1
Solution:
d
(sin(x)
dx
+ 4cos(3y)) =
d
(1)
dx
cos(x) − 4sin(3y) · 3y ′ = 0
cos(x) − 12sin(3y)y ′ = 0
−12cos(3y)y ′ = −cos(x)
−cos(x)
cos(x)
y ′ = −12sin(3y) = 12sin(3y)
2.5.4
Finding the Derivatives of the inverse Trigonometric Functions Using Implicit Differentiation
Example 11: The Derivative of y=arccsin(u)
Let y = arcsin(u)
If y = arcsin(u) ,then u = sin(y)
Now, find the implicit derivative.
d
(u)
dx
d
= dx
(sin(y))
(1) · u′ = cos(y)y ′
u′ = cos(y)y ′
′
u
y ′ = cos(y)
Recall the following Pythagorean Identity:
sin2 y + cos2 y = 1
cos2 y = 1 −√sin2 y
√
cos2 y =√ 1 − sin2 y
cos(y) = 1 − sin2 y
′
′
′
u
u
Thus, y ′ = cos(y)
= √ u 2 = √1−u
2
1−sin y
30
Example 12: The Derivative of y=arccos(u)
Let y = arccos(u)
If y = arccos(u) ,then u = cos(y)
Now, find the implicit derivative.
d
(u)
dx
d
= dx
(cos(y))
′
(1) · u = −sin(y)y ′
u′ = −sin(y)y ′
′
u
y ′ = − sin(y)
Recall the following Pythagorean Identity:
sin2 y + cos2 y = 1
sin2 y = 1 −√cos2 y
√
sin2 y =√ 1 − cos2 y
sin(y) = 1 − cos2 y
′
′
′
u
u
Thus, y ′ = − sin(y)
= − √ u 2 = − √1−u
2
1−cos y
Example 13: The Derivative of y=arctan(u)
Let y = arctan(u)
If y = arctan(u) ,then u = tan(y)
Now, find the implicit derivative.
d
(u)
dx
d
= dx
(tan(y))
(1) · u′ = tan(y)y ′
u′ = sec2 yy ′
′
y ′ = secu2 (y)
Recall the following Pythagorean Identity:
1 + tan2 y = sec2 y
′
′
′
u
u
y ′ = secu2 (y) = 1+tan
2 y = 1+u2
31
Summary of the derivatives of the inverse trigonmetric functions.
′
d
(arcsin(u))
dx
= √u
d
(arccos(u))
dx
= −√ u
d
(arctan(u))
dx
u
= 1+u
2
1−u2
′
1−u2
′
32
2.6
2.6.1
Inverse Functions
The Inverse Trigonometric Functions
In the previous section, we derived the derivatives of the arcsin(x), arccos(x), and arctan(x).
After deriving the derivatives of the first three arc functions, let look at the derivatives of
all six inverse trigonometric functions.
′
d
(arcsin(u))
dx
= √u
d
(arccos(u))
dx
= −√ u
d
(arctan(u))
dx
u
= 1+u
2
1−u2
′
1−u2
′
′
d
(arccot(u))
dx
u
= − 1+u
2
d
(arcsec(u))
dx
=
d
(arccsc(u))
dx
=−
2.6.2
′
√u
|u|· u2 −1
′
√u
|u|· u2 −1
Finding the Derivatives of the Inverse Trigonometric Functions
Example 1
Find the derivative of y = x2 + arccos(x)
Solution:
y = x2 + arccos(x)
x′
y ′ = 2x − √1−x
2
1
′
√
y = 2x − 1−x2
Example 2
Find the derivative of y = arccos(3x2 )
Solution:
y = arccos(3x2 )
y′ = − √
(3x2 )′
1−(3x2 )2
6x
= − √1−9x
4
Example 3
Find the derivative of y = arcsin(3x)
33
Solution:
y = arcsin(3x)
(3x)
y′ = √
′
=
1−(3x)2
√ 3
1−9x2
Example 4
Find the derivative of f (x) = x3 arcsin(x)
Solution:
To find the derivative, we need to use the product rule.
f (x) = x3 arcsin(x)
f ′ (x) = (arcsin(x))′ x3 + (x3 )′ arcsin(x)
′
f ′ (x) = √ (x) 2 · x3 + (x3 )′ arcsin(x)
=
3
√x
1−x2
1−(x)
2
+ 3x arcsin(x)
Example 5
Find the derivative of f (x) = 2x2 arctan(x)
Solution:
f (x) = 2x2 arctan(x)
f ′ (x) = (arctan(x))′ (2x2 ) + (2x2 )′ · arctan(x)
(x)′
f ′ (x) = 1+x2 ·2x2 + (4x)arctan(x)
2x2
f ′ (x) = 1+x
2 +(4x)arctan(x)
Example 6
Find the slope of a tangent line to the function f (x) = 2arccos(x) at the point ( 21 , 2π
)
3
Solution:
Find the derivative first.
′
x
√ 2
f ′ (x) = 2[− √1−x
2 ] = − 1−x2
Now, substitute
f ′ (x) = − √ 2 1
1
2
1−( 2 )2
into the derivative.
√
= − √ 2 1 = − √2 3 = − √43 = − 4 3 3
1− 4
4
34
2.7
Related Rates
Related rates problems involve using implicit derivatives to solve for quantities that are rates
of change that are dependent on time. Each of these rates will involve the implicit derivative
term such as dy
or dx
.
dt
dt
2.7.1
Applications of Related Rates
Now, let’s look at an example where volume is drained from a cone shaped container. When
water is drained out of a conical tank, the variables for volume, radius, and height of the
water level are all functions of time. We will start by using the basic formula for the volume
of a cone.
V =
π
3
· r2 h
If you differentiate the above equations you will get the following related rates equation
dV
dt
dV
dt
dV
dt
dV
dt
= dtd [ π3 · r2 h]
= π3 · [(h)′ r2 + (r2 )′ h]
= π3 · [r2 · dh
+ 2rh dr
]
dt
dt
π 2 dh
2π
dr
= 3 r · dt + 3 rh · dt
Example 1
Assume that x and y are both differentiable functions of t and find the requires values of
and dx
.
dt
Equation
1) x · y = 3
2) x2 + y 2 = 5
Find
when x = 1
x = 1 and y = 2
dy
dx
Given
=2
=2
dx
dt
dx
dt
Solution:
Part 1
First, find the value of y.
xy = 3
(1) · y = 3
y=3
Now, find the implicit derivative.
= dtd (3)
(y) x + (x)′ y = 0
+ y · dx
=0
x · dy
dt
dt
d
(xy)
dt
′
Next, solve for
dy
dt
35
dx
dt
1·
dy
dt
dy
dt
+3·2=0
+6=0
= −6
dy
dt
Part 2
Find the implicit derivative
x2 + y 2 = 5
d
(x2 + y 2 ) = dtd (5)
dt
2x · dx
+ 2y · dy
=0
dt
dt
Now, use substitution to find the value of
2(1)(2) + 2(2) ·
4 + 4 · dy
=0
dt
dy
4 · dt = −4
dy
= −1
dt
dy
dt
dy
dt
=0
Example 2
The area of circle is increasing at a rate of 4 centimeters per minute. Find the rate of change
of the area when a) r = 5 centimeters b) r = 12 centimeters
Solution:
Part a: r = 5
dr
dt
=4
A = π · r2
dA
= dtd (πr2 )
dt
dA
= 2πr · dr
dt
dt
dA
=
2π
·
5
·4
dt
dA
cm2
= 40π s
dt
36
Part b: r = 12
dr
dt
=4
A = π · r2
dA
= dtd (πr2 )
dt
dA
= 2πr · dr
dt
dt
dA
=
2π
·
12
·4
dt
dA
cm2
= 96π s
dt
Example 3
An airplane flying at an altitude of 4 miles passes directly over a radar tracking station, as
shown below. When the plane is 5 miles away (s = 5), the radar detects that the distance s
is changing at a rate of 440 miles per hour. What is the speed of the plane?
Solution:
Use the Pythagorean Theorem to find the value of x.
s2 = x2 + y 2
52 = x2 + 42
25 = x2 + 16
25 − 16 = x2 + 16 − 16
2
x
√ =9√
x2 = 9
x=3
37
Using the Pythagorean Theorem again, we get the following equation.
s2 = x2 + 42
s2 = x2 + 16
Now, differentiate the equation implicitly.
d
(s2 ) = dtd (x2 + 16)
dt
2s · ds
= 2x · dx
dt
dt
= 440, s = 5, and x = 3
Next, substitute ds
dt
dx
2(5)(440) = 2(3) · dt
4400 = 6 · dx
dt
dx
4400
=
=
733 miles per hour
dt
6
Example 4
Find the rate of change of the volume of a cone, if
when r = 3 inches.
dr
dt
is 2 inches per minute and h = 3r
Solution:
Since h = 3r ⇒
dh
dt
=3·
dr
dt
= 3 · 2 = 6 inches
h = 3r = 3(3) = 9 inches
Recall that the volume of a cone is V = 13 πr2 h
V = 13 πr2 h
d
(V ) = dtd ( 13 πr2 h)
dt
dV
= 13 π(h)′ r2 + 13 π(r2 )′ h
dt
dV
= 13 πr2 dh
+ 23 πrh dr
dt
dt
dt
dV
1
2
2
=
π(3)
·
6
+
π(3)(9)(2)
dt
3
3
dV
= 13 π(54) + 23 π(54)
dt
dV
= 54π in3
dt
Example 5
The radius of a sphere is increasing at a rate 2 inches per minute. find the rate of change of
the volume when r = 4 inches and r = 10 inches.
Solution:
Given
dr
dt
=2
Recall that V = 43 πr3
V = 34 πr3
d
V = dtd ( 43 πr3 )
dt
38
dV
dt
dV
dt
= 43 π(3r2 dr
)
dt
2 dr
= 4πr · dt
When r = 4 inches ⇒
dV
dt
When r = 10 inches ⇒
dV
dt
= 4πr2 ·
dr
dt
= 4πr2 ·
= 4π(4)2 = 64π
dr
dt
= 4π(10)2 = 400π
39