DIFFERENTIATION
2.2 The limit of a function
Examp1e 1
Determine
1. lim 2x 2 − 3 = 5
x→2
2. lim 2y 2 − 3 = 2y 2 − 3
x→2
sin x
=1
x→0 x
4. lim (1 + x)1/x = 2.71828... = e
3. lim
x→0
2.3 Continuity
Examp1e 2
Let f , g and h be functions defined by
f (x) = cos x − x 5 ,
√
g (x) = x x,
h(x) =
sin x
.
x
Answer the following and motivate your answer.
1. Is f continuous in 7? Yes.
2. Is g continuous in 4? Yes.
3. Is g continuous in 0? No, only right continuous, since g
is not defined for negative real numbers.
4. Is g continuous in −3? No, g (−3) does not exist.
5. Why is h not continuous in 0? Can we extend the
definition of h in order to be continuous every where?
h(0) does not exist. Define h(0) = 1.
2.4 Rate of change
Page 57: Average rate of change is an approximation of the
instantaneous rate of change. Decreasing on the magnitude of
h increases on the accuracy of the instantaneous rate of
change. Take note that h may attain negative values as well.
2.5 The derivative of a function
Page 59: Four steps in finding derivatives (instantaneous rate
of change)
Example 3
Find f ′ (4) if f (x) = 2x − x 2 . f ′ (4) = −6.
Example 4
Find f ′ (x) if f (x) = 2x − x 2 . f ′ (x) = 2 − 2x.
Example textbook p 62
2.6 Rules of differentiation
Study Rules 1 to 6 in the textbook.
Rule 7: The chain rule
Example 5
Evaluate f g (x) , f ′ (x), g ′ (x) and f ′ g (x) if
1. f (x) = 2x and g (x)= x 2 . f g (x) = 2x 2 , f ′ (x) = 2,
g ′ (x) = 2x, f ′ g (x) = 4x.
2. f (x) = x 2 and g (x) = 2x. f g (x) = 4x 2 , f ′ (x) = 2x,
g ′ (x) = 2, f ′ g (x) = 8x.
3. f (x) = x 2 and g (x) = 2x − 6. f g (x) = 4x 2 − 24x + 36,
f ′ (x) = 2x, g ′ (x) = 2, f ′ g (x) = 8x − 24.
√
√
4. f (x) = x and g (x) = x 3 − 1. f g (x) = x 3 − 1,
f ′ (x) = 0.5x −0.5, g ′ (x) = 3x 2 . How do we calculate
f ′ g (x) ? Here we have to use the chain rule.
Example 6
Determine f ′ (x) for each of the questions in Example 5 using
the chain rule. Nos. 1 – 3 are for homework. No. 4: Set
√
dy
dy du
u = x 3 − 1 and y = f (u) = u. Then
=
·
=
dx
du dx
−0.5
0.5u −0.5 · 3x 2 = 1.5x 2 (x 3 − 1) .
Exercise 10 page 74
dy
= (x 2 − 7) (3x 2 + 10x + 9) + 2x (x 3 + 5x 2 + 9x)
dx
Exercises 17 page 74
Set g (x) = (1 + x) (1 + 2x − 3x 2 ) and h(x) = x − 4.
g (x)
dy
g ′ (x)h(x) − g (x)h′ (x)
Then y =
and
=
=
2
h(x)
dx
h(x)
h
i
2
(1+x) (2−6x)+(1+2x −3x ) x − 4 − (1 + x) (1 + 2x − 3x 2)
2
x −4
2.7 Inverse functions and their derivatives
Consider a one-to-one function f with domain [a, b] and range
[c, d ]. Then the inverse function f −1 exists and
y = f (x) ⇐⇒ x = f −1 (y ).
The graphs of f and f −1 are symmetric about the line y = x.
Example 7
Find the inverse function of f (x) = 2x − 4 and draw both
df −1
df
and
.
functions on the same system of axes. Find
dx
dx
Set y = 2x − 4 (= f (x)) and write x in terms of y to find
dx
dy
= 2 and
= 0.5.
x = 0.5(y + 4) = f −1 (y ). Then
dx
dy
dy
1
df
Note that
=
and that
= 2. Reversing the
dx
(dx/dy )
dx
role of x and y in x = 0.5(y + 4) = f −1 (y ), we may write
y = 0.5(x + 4) = f −1 (x) and hence df −1 /dx = 0.5.
Example 8
Find the inverse function of f (x) = x 2 − 4, where x ∈ (0, ∞)
and draw both functions on the same system of axes.
df −1
df
and
.
Find
dx
dx
Note that the domain of f ensures a one-to-one function.
Set y = x 2 − 4 and write x in terms of y to find
√
dy
x = y + 4 = f −1 (y ). Then
= 2x and
dx
dx
= 0.5(y + 4)−0.5 . Note that
dy
dy
1
(dx/dy ) = 0.5/x = 1/(2x) and hence
=
.
dx
(dx/dy )
df
Furthermore,
= 2x. Reversing the role of x and y in
dx
√
√
x = y + 4 = f −1 (y ), we get y = x + 4 = f −1 (x)) and
hence df −1 /dx = 0.5(x + 4)−0.5 .
2.8 Derivatives of special functions
Alternative for Rules 9 and 10
d x
(e ) = e x
Rule 9(a):
dx
Rule 9(b): Note that ax = e x ln a . The chain rule yields
d
d x
e x ln a
(a ) =
dx
dx
d
= e x ln a (x ln a)
dx
x ln a
= e
(ln a)
x
= a ln a
Rule 10(a):
Rule 10(b):
d
e f (x) = e f (x) f ′ (x)
dx
d
af (x) = af (x) f ′ (x) ln a
dx
Rule 11(a):
d
1
ln x =
dx
x
ln x
. The chain rule yields
ln a
d
d
1
(loga x) =
ln x
dx
ln a
dx
1
1
=
ln a
x
Rule 11(b):
Rule 12(a):
Rule 12(b):
Note that loga x =
f ′ (x)
d
ln f (x) =
dx
f (x)
′ f (x)
1
d
loga f (x) =
dx
ln a
f (x)
2.9 Higher order derivatives
Example 9
Find the following derivatives.
1. f (x) = e 2x , f ′′′ (x) = 8e 2x
2 − 2 ln v
2. g (v ) = (ln v )2 , g ′′ (v ) =
v2
12
3. h(y ) = ln y 2, h(4) (y ) = − 4
y
d
2
4. z(t) = e t log10 t,
z(t) = 0
dx
Example 10
A 350 kg motorcycle moves on a straight road according to
the relationship x = 1.4t 3 + 5t, where x denotes distance in m
and t time in s.
1. Calculate its speed after 3 seconds.
2. If it hits a concrete wall after 3 seconds, what would the
impact force be on the wall?
Answer
dx
= 4.2t 2 + 5, hence at time t = 3 s, the speed is
dt
v (3) = 42.8 m/s (≈ 154 km/h).
dv
d 2x
2. a =
= 2 = 8.4t, hence a(3) = 25.2 m/s 2 and the
dt
dt
impact force is F = ma = 8.82 kN.
1. v =
2.10 Optimisation
2.10.1 Extreme values for single variable functions
The critical values of a continuous function f defined on [a, e]
are found
1. at the endpoints a and e,
2. at points p where f ′ (p) = 0 and
3. at points q where f ′ (q) is not defined.
Refer to Figure 2.10.1 p 89; points b and d are typical for
point p and d is a typical example for point q.
Remarks
◮
◮
◮
The local maxima and minima (extreme values) form part
of the list of critical values.
Not all critical values are local extrema.
The absolute extremes form part of the list of local
extrema.
Suppose c is a critical point of f with f ′ (c) = 0 and suppose
f ′′ (c) exists.
1. If f ′′ (c) < 0, f (c) is a local maximum.
2. If f ′′ (c) > 0, f (c) is a local minimum.
3. If f ′′ (c) = 0, no conclusion can be drawn, and the
behaviour of f ′′ in the point c must be investigated. The
value f (c) may be
◮
◮
◮
a local minimum or
a local maximum or
not any one of the above mentioned options.
Example 11
Find the absolute minimum and maximum of the function f ,
with f (x) = 2x 3 − x 2 − 2x + 1 and −1.2 ≤ x ≤ 1.2.
1. Step 1 Find all critical values
1.1 The values f (−1.2) = −1.496 and f (1.2) = 0.616 are
critical values.
1.2 f ′ (x) = 0 ⇐⇒ 6x 2 − 2x − 2 = 0 ⇐⇒ x = 0.76759 or
x = −0.43426 and the corresponding critical values are
f (0.76759) = −0.21985 and f (−0.43426) = 1.51615.
1.3 There are no points q for which f ′ (q) does not exist.
2. Step 2 Identify local / relative extrema
f ′′ (x) = 12x − 2 and
2.1 f ′′ (0.76759) = 7.21110 > 0 =⇒
f (0.76759) = −0.21985 is a local / relative minimum.
2.2 f ′′ (−0.43426) = −7.21110 < 0 =⇒
f (−0.43426) = 1.51615 is a local / relative maximum.
3. Step 3 Identify absolute extrema
The absolute maximum is f (−0.43426) = 1.51615 and
the absolute minimum is f (−1.2) = −1.496.