MTH4100 Calculus I
Lecture notes for Week 5
Thomas’ Calculus, Sections 2.6 to 3.5 except 3.3
Rainer Klages
School of Mathematical Sciences
Queen Mary University of London
Autumn 2009
Continuous extension to a point
example: sin x
f (x) =
x
sin x
is defined and continuous for all x 6= 0. As lim = 1, it makes sense to define a new
x→0 x
function sin x
for x 6= 0
F (x) = x
1 for x = 0
Definition 1 If lim f (x) = L exists, but f (c) is not defined, we define a new function
x→c
f (x) for x 6= c
F (x) =
L for x = c ,
which is continuous at c. It is called the continuous extension of f (x) to c.
A function has the intermediate value property if whenever it takes on two values, it also
takes on all the values in between.
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Geometrical interpretation of this theorem: Any horizontal line crossing the y-axis be-
tween f (a) and f (b) will cross the curve y = f (x) at least once over the interval [a, b].
Continuity is essential: if f is discontinuous at any point of the interval, then the function
may “jump” and miss some values.
Differentiation
Recall our discussion of average and instantaneous rates of change.
example: revisit growth of fruit fly population
basic idea:
• Investigate the limit of the secant slopes as Q approaches P .
• Take it to be the slope of the tangent at P .
Now we can use limits to make this idea precise. . .
example: Find the slope of the parabola y = x2 at the point P (2, 4).
• Choose a point Q a horizontal distance h 6= 0 away from P ,
Q(2 + h, (2 + h)2 ) .
• The secant through P and Q has the slope
∆y (2 + h)2 − 22 4 + 4h + h2 − 4
= = =4+h.
∆x (2 + h) − 2 h
• As Q approaches P h approaches 0, hence
∆y
m = lim = lim (4 + h) = 4
h→0 ∆x h→0
must be the parabola’s slope at P .
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• The equation of the tangent through P (2, 4) is y = y1 + m(x − x1 );
here: y = 4 + 4(x − 2) or y = 4x − 4.
summary:
Now generalise to arbitrary curves and arbitrary points:
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example: Find slope and tangent to y = 1/x at x0 = a 6= 0
1 1
1. f (a) = , f (a + h) =
a a+h
2. slope: f (a + h) − f (a)
m = lim
h→0 h
1 1
−
= lim a+h a
h→0 h
a − (a + h)
= lim
h→0 h · a(a + h)
−1 1
= lim =− 2
h→0 a(a + h) a
1 1 2 x
3. tangent line at (a, 1/a): y = + − 2 (x − a) or y = − 2 .
a a a a
f (x0 + h) − f (x0 )
The expression is called the difference quotient of f at x0 with incre-
h
ment h. The limit as h approaches 0, if it exists, is called the derivative of f at x0 .
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Let x ∈ D(f ).
If f ′ (x) exists, we say that f is differentiable at x.
Choose z = x + h: h = z − x approaches 0 if and only if z → x.
d dy
Equivalent notation: If y = f (x), y ′ = f ′ (x) = f (x) = .
dx dx
Calculating a derivative is called differentiation (“derivation” is something else!).
x
example: Differentiate from first principles f (x) = .
x−1
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f (x + h) − f (x)
f ′ (x) = lim
h→0 h
x+h x
x+h−1
− x−1
= lim
h→0 h
1 (x + h)(x − 1) − x(x + h − 1)
= lim
h→0 h (x + h − 1)(x − 1)
1 −h
= lim
h→0 h (x + h − 1)(x − 1)
1
= −
(x − 1)2
√
example: Differentiate f (x) = x by using the alternative formula for derivatives.
f (z) − f (x)
f ′ (x) = lim
z→x z−x
√ √
z− x
= lim
z→x z−x
√ √
z− x
= lim √ √ √ √
z→x ( z − x)( z + x)
1
= lim √ √
z→x z+ x
1
= √
2 x
note: For f ′ (x) at x = 4, one sometimes writes
d √
′ 1
f (4) = x = √ .
dx x=4 2 x x=4
One-sided derivatives
In analogy to one-sided limits, we define one-sided derivatives:
f (x + h) − f (x)
lim+ right-hand derivative at x
h→0 h
f (x + h) − f (x)
lim− left-hand derivative at x
h→0 h
f is differentiable at x if and only if these two limits exist and are equal.
example: Show that f (x) = |x| is not differentiable at x = 0. [2009 exam question]
• right-hand derivative at x = 0:
|0 + h| − |0| |h|
lim+ = lim+ =1
h→0 h h→0 h
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• left-hand derivative at x = 0:
|0 + h| − |0| |h|
lim− = lim− = −1 ,
h→0 h h→0 h
so the right-hand and left-hand derivatives differ.
Theorem 1 If f has a derivative at x = c, then f is continuous at x = c.
Proof: Trick: For h 6= 0, write
f (c + h) − f (c)
f (c + h) = f (c) + h.
h
f (c + h) − f (c)
By assumption, lim = f ′ (c). Therefore,
h→0 h
lim f (c + h) = f (c) + f ′ (c) · 0 = f (c) .
h→0
According to definition of continuity, f is continuous at x = c. q.e.d.
caution: The converse of the theorem is false!
note: The theorem implies that if a function is discontinuous at x = c, then it is not dif-
ferentiable there.
Differentiation rules (‘machinery’)
Proof of one rule see ff; proof of other rules see book, Section 3.2.
Rule 1 (Derivative of a Constant Function) If f has the constant value f (x) = c,
then
df d
= (c) = 0 .
dx dx
Rule 2 (Power Rule for Positive Integers) If n is a positive integer, then
d n
x = nxn−1 .
dx
Rule 3 (Constant Multiple Rule) If u is a differentiable function of x, and c is a con-
stant, then
d du
(cu) = c .
dx dx
Proof:
d
cu =
dx
cu(x + h) − cu(x)
(def. of derivative) = lim
h→0 h
u(x + h) − u(x)
(limit laws) = c lim
h→0 h
du
(u is differentiable) = c
dx
q.e.d.
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Rule 4 (Derivative Sum Rule) If u and v are differentiable functions of x, then
d du dv
(u + v) = + .
dx dx dx
example: Differentiate y = 3x4 + 2.
dy d
= (3x4 + 2)
dx dx
d d
(rule 4) = (3x4 ) + (2)
dx dx
d 4 d
(rule 3) = 3 (x ) + (2)
dx dx
d
(rule 2) = 3 · 4x3 + (2)
dx
(rule 1) = 12x + 0 = 12x3
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Rule 5 (Derivative Product Rule) If u and v are differentiable functions of x, then
d du dv
(uv) = v+u .
dx dx dx
example: Differentiate y = (x2 + 1)(x3 + 3).
here: u = x2 + 1 , v = x3 + 3
u′ = 2x , v ′ = 3x2
y ′ = 2x(x3 + 3) + (x2 + 1)3x2 = 5x4 + 3x2 + 6x
Rule 6 (Derivative Quotient Rule) If u and v are differentiable functions of x and
v(x) 6= 0, then
du dv
d u dx
v − u dx
= .
dx v v2
t−2
example: Differentiate y = .
t2 + 1
here: u = t − 2 , v = t2 + 1
u′ = 1 , v ′ = 2t
1(t2 + 1) − (t − 2)2t −t2 + 4t + 1
y′ = =
(t2 + 1)2 (t2 + 1)2
Common mistakes: (uv)′ = u′ v ′ and (u/v)′ = u′ /v ′ are generally WRONG!
Rule 7 (Power Rule for Negative Integers) If n is a negative integer and x 6= 0, then
d n
x = nxn−1 .
dx
(Proof: define n = −m and use the quotient rule.)
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d 1 d −11
example: = (x ) = −11x−12 .
dx x11 dx
Higher-order derivatives
If f ′ is differentiable, we call f′′ =(f ′ )′ the second derivative of f .
d2 y d dy dy ′
Notation: f ′′ (x) = 2 = = = y ′′ .
dx dx dx dx
Similarly, we write f ′′′ = (f ′′ )′ for the third derivative, and generally for the n-th deriva-
tive, n ∈ N0 : f (n) = (f (n−1) )′ with f (0) = f .
example: Differentiate repeatedly f (x) = x3 and g(x) = x−2 .
f ′ (x) = 3x2 g ′ (x) = −2x−3
f ′′ (x) = 6x g ′′ (x) = 6x−4
f ′′′ (x) = 6 g ′′′ (x) = −24x−5
f (4) (x) = 0 g (4) (x) = . . .
