The exact change in x and y is ∆x and ∆y , respectively.
The differentials dx and dy are related through dy = f 0 (x ) dx .
The exact change and differentials have the following relationship:
∆x = dx and ∆y ≈ dy
AMAT 217 (University of Calgary)
Fall 2013
1/9
Example
Let y = (x + 1)2 . Find the exact change in x and y , along with the differentials dx and
dy , if x changes from 3 to 3.5.
The exact change in x is:
∆x = 3.5 − 3
→
∆x = 0.5
The exact change in y is:
∆y
=
(3.5 + 1)2 − (3 + 1)2
=
(4.5)2 − 42
=
20.25 − 16
=
4.25
The differential dx is: dx = ∆x = 0.5.
The differential dy can be obtained from the derivative:
dy
= 2(x + 1)
dx
Using x = 3 and dx = 0.5 we get:
dy = 2(3 + 1)(0.5)
→
→
dy = 2(x + 1) dx
dy = 8(0.5)
→
dy = 4
Note that ∆y ≈ dy , so dy would provide a good approximation to ∆y in this case.
AMAT 217 (University of Calgary)
Fall 2013
2/9
Relative and percentage change
The difference between two values is not always a good comparison metric.
Example: Giving $1,000,000 to Bill Gates would not be very relevant to his
net worth, whereas, giving $1,000,000 to Mike is very relevant to his net
worth (since Mike is poor). Although the change in net worth is the same for
both individuals ($1,000,000 increase), a better comparison is to use the
relative (or percentage) change of net worth.
The relative change takes into account the “size” of the quantities involved
and is defined in words as:
absolute change
relative change =
reference value
Example: The relative change when moving from 5 to 6 is |6−5|
= 0.2 (or a
5
20% increase) whereas the relative change when moving from 1000 to 1001 is
|1001−1000|
= 0.001 (or a 0.1% increase).
1000
Definition: Relative and Percentage Change
The relative change in z is
∆z
∆z
× 100%.
. The percentage change in z is
z
z
* This is sometimes also called the relative (or percentage) error.
AMAT 217 (University of Calgary)
Fall 2013
3/9
Example
By approximately what percent does the area of a circle increase if the radius increases by 2%?
We are given the percentage change in radius (r ) to be 2% (relative change is 0.02):
∆r
relative change in r =
= 0.02
Given information
r
The question asks for the percentage change in area (A), that is, we need to compute:
∆A
percentage change in A =
× 100%
A
It is difficult to compute ∆A exactly in this case, so we use differentials:
dA
∆A
≈
∆r = dr
and
∆A ≈ dA,
that is,
∆r
dr
We have A = πr 2 , so
∆A
dA
≈ 2πr
→
∆A ≈ 2πr ∆r
= 2πr
→
∆r
dr
∆A
We want to determine an approximation for A , so we divide both sides by A:
∆A
2πr ∆r
2πr ∆r
∆r
≈
=
=2
A
A
πr 2
r
Since
∆r
r
= 0.02, we have
∆A
≈ 2(0.02) = 0.04
A
The percentage change in area is an approximate increase of 4%.
AMAT 217 (University of Calgary)
Fall 2013
4/9
Example
Find f (46334665)
p
23
123! sin(1)
r
if f (x ) = x 21372
144 sin8 (1)
9π 4
72π 2 sin4 (1)
+ 42740 +
.
42744
x
x
x 42742
***SIMPLIFY FIRST!*** Recall that we can factor perfect square trinomials as:
a2 + 2ab + b 2 = (a + b)2
and
a2 − 2ab + b 2 = (a − b)2
In our question, it’d be nice if we had a perfect square under the square root.
We first arrange the terms from highest to lowest power (−42740 is bigger than −42742):
9π 4
72π 2 sin4 (1)
144 sin8 (1)
(highest exponent),
,
(lowest exponent)
x 42744
x 42740
x 42742
Use the (positive) square root of the highest and lowest terms for a and b:
3π 2
12 sin4 (1)
a = 21370
and
b=
x
x 21372
Then take the sign of the middle term, in this case positive, and factor appropriately:
9π 4
72π 2 sin4 (1)
144 sin8 (1)
+
+
42740
42742
x
x
x42744
? =?
3π 2
12 sin4 (1)
+
21370
x
x21372
Expanding we can check that indeed we have that +2ab equals the middle term
Thus, the factoring works and we can simplify f (x ) to:
s
f (x ) = x
21372
3π 2
x 21370
+
12 sin4 (1)
x 21372
2
3π2
= x 21372 x 21370
2
72π 2 sin4 (1)
.
x 42742
+
12 sin4 (1) = 3π 2 x 2 +12 sin4 (1)
x 21372 p
f 0 (x ) = 6π 2 x , f 00 (x ) = 6π 2 and f (n) (x ) = 0 for n ≥ 3, so, f (46334665) ( 23 123! sin(1)) = 0.
AMAT 217 (University of Calgary)
Fall 2013
5/9
Example
How many tangent lines to the function f (x ) =
x
pass through the point (1, 2)?
x +1
*Note that the point (1, 2) does NOT lie on the function f (x ).
Then, the tangent line to f is at some unknown point x = a with y -coord
The slope of this straight line through points (1, 2) and
m=
a
−2
RISE
= a+1
=
RUN
a−1
a
a+1
a,
a
a+1
a
.
a+1
is:
−a−2
− 2(a+1)
−(a + 2)
a+1
= a+1 =
a−1
a−1
(a + 1)(a − 1)
But the derivative at x = a is equal to the slope of the tangent line at x = a.
The derivative of f (x ) using the quotient rule is:
f 0 (x ) =
(1)(x + 1) − (x )(1)
1
=
(x + 1)2
(x + 1)2
→
m = f 0 (a) =
1
(a + 1)2
Setting the two equations for slope equal we get one equation with an unknown a:
−(a + 2)
1
=
(a + 1)(a − 1)
(a + 1)2
→
− (a + 2)(a + 1) = (a − 1)
→
This quadratic equation has two solutions for a, namely, a = −2 ±
are two tangent lines to f (x ) that go through the point (1, 2).
AMAT 217 (University of Calgary)
a2 + 4a + 1 = 0
√
3, thus, there
Fall 2013
6/9
Example
Suppose f (x ) and g(x ) are differentiable functions such that f (g(x )) = x and
f 0 (x ) = 1 + (f (x ))2 . Find a formula for g 0 (x ) as a function of x only.
We start by trying to differentiate both sides of f (g(x )) = x .
We use the chain rule [f (g(x ))]0 = f 0 (g(x )) · g 0 (x ) on the left side and the
derivative formula (ax )0 = a with a = 1 for the right side to get:
f 0 (g(x)) · g 0 (x ) = 1 (∗)
Note we are given that f 0 (x ) = 1 + (f (x ))2 , so replacing x with g(x ) we get
a formula for f 0 (g(x)):
f 0 (g(x)) = 1 + (f (g(x )))2
Substituting this into (∗) gives:
1 + (f (g(x )))2 · g 0 (x ) = 1
Since f (g(x )) = x we can simplify this to:
(1 + x 2 ) · g 0 (x ) = 1
AMAT 217 (University of Calgary)
→
g 0 (x ) =
1
1 + x2
Fall 2013
7/9
An extra limit technique
Example
Compute the following limit: limπ
θ→ 3
cos θ − 0.5
.
θ − π/3
Note that cos(π/3) = 0.5 so this is a 0/0 limit type.
Currently, none of our limit techniques will work for this limit, but we can
recognize this limit as the derivative of a certain function!
Recall the algebraic definitions of a derivative:
f (x ) − f (a)
f (x + h) − f (x )
and
f 0 (a) = lim
x →a
h→0
h
x −a
The second formula seems to match our problem best using θ in place of x .
It appears that f (θ) = cos θ and a = π/3 will work since f (π/3) = 0.5.
Then we have:
cos θ − 0.5
f 0 (π/3) = limπ
using f (θ) = cos θ and a = π/3
θ→ 3
θ − π/3
√
Since f 0 (θ) = − sin θ, we have f 0 (π/3) =
√ − sin(π/3) = − 3/2.
cos θ − 0.5
3
Thus, the limit is limπ
=−
.
θ→ 3
θ − π/3
2
f 0 (x ) = lim
AMAT 217 (University of Calgary)
Fall 2013
8/9
Example
Find constants a, b so that f (x ) =
ax 2 + 2x
4b
if x < 2
if x ≥ 2
is differentiable at x = 2.
For differentiability at x = 2 we need f (x ) to:
(i) be continuous at x = 2,
[Recall: differentiable → continuous but continuous 6→ differentiable]
(ii) have matching right and left derivatives at x = 2.
The first condition requires that left and right hand limits be equal:
lim f (x ) = lim (ax 2 + 2x ) = 4a + 4
x →2−
x →2−
and
lim f (x ) = lim (4b) = 4b
x →2+
x →2+
Thus, we require 4a + 4 = 4b, that is, a + 1 = b.
The second condition requires that left and right derivatives are equal:
f−0 (2) =
d
= (2ax + 2) = 4a + 2
ax 2 + 2x dx
x =2
x =2
and
f+0 (2) =
d
=0
(4b) dx
x =2
Thus, we require 4a + 2 = 0, that is, 2a + 1 = 0.
We have two equations and two unknowns: a + 1 = b and 2a + 1 = 0.
Solving this gives a = −1/2 and b = 1/2.
Therefore, for f (x ) to be differentiable at x = 2 we require a = −1/2 and b = 1/2.
AMAT 217 (University of Calgary)
Fall 2013
9/9