University of Zimbabwe
SEMESTER 1: 2020
Lecturer : Mr. T. M. Mazikana
Course title : Mathematical Methods for Physics/Meteorology/Forensic Physics 1
Course code :
Section 3: Differentiation and Integration
November 4, 2020
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University of Zimbabwe
SEMESTER 1: 2020
Lecturer : Mr. T. M. Mazikana Course title : Mathematical Methods for Physics/Meteorology/Forensic Physics 1 Course code :
Section 3: Differentiation and Integration
November 4, 2020
Chapter 1
Differentiation
Increments. The increment ∆x of a variable x is the change in x as it increases or decreases from one value x = x 0 to another value x = x 1 in its domain. Here, ∆x = x 1 − x 0 and we may write x 1 = x 0 + ∆x. If the variable x is given an increment ∆x from x = x 0 (i.e., if x changes from x = x 0 to x 1 = x 0 + ∆x) and a function y = f (x) is thereby given an increment ∆y = f (x 0 + ∆x) − f (x 0 ) from y = f (x 0 ), then the quotient ∆y ∆x
change in y change in x
is called the average rate of change of the function on the interval between x = x 0 and x 1 = x 0 +∆x.
Let f (x) be defined at any point x 0 in (a, b). The derivative of f (x) at x = x 0 is defined as
f ′(x 0 ) = lim h→ 0 f (x 0 + h) − f (x 0 ) h if this limit exists. A function is called differentiable at a point x = x 0 , if it has a derivative at that point, i.e., if f ′(x 0 ) exists. If we write x = x 0 + h, then h = x − x 0 and h approaches 0 if and only if x approaches x 0. Therefore, an equivalent way of stating the definition of the derivative, is
f ′(x 0 ) = lim x→x 0
f (x) − f (x 0 ) x − x 0
Example: If f (x) = x^3 − x, find a formula for f ′(x).
Solution:
f ′(x) = lim h→ 0
f (x + h) − f (x) h = lim h→ 0
[(x + h)^3 − (x + h)] − [x^3 − x] h = lim h→ 0
x^3 + 3x^2 h + 3xh^2 + h^3 − x − h − x^3 + x h = (^) hlim→ 0
3 x^2 h + 3xh^2 + h^3 − h h = lim h→ 0 (3x^2 + 3xh + h^2 − 1) = 3x^2 − 1.
For x = 0 we have to investigate
f ′(0) = (^) hlim→ 0 f (0 + h) − f (0) h = (^) hlim→ 0 |0 + h| − | 0 | h (if it exists).
Let’s compute the left and right limits separately;
lim h→ 0 +
|0 + h| − | 0 | h = lim h→ 0 +
|h| h = lim h→ 0 +^
lim h→ 0 −
|0 + h| − | 0 | h
= lim h→ 0 −
|h| h
= lim h→ 0 −
Since these limits are different f ′(0) does not exist. Thus, f is differentiable at all x except 0.
1.1 Differentiation Techniques (Finding Derivatives)
f ′(x 0 ) = lim h→ 0 f (x 0 + h) − f (x 0 ) h if this limit exists.
- The derivative of any constant function is zero, i.e., c′^ = 0.
- For any real number n, (xn)′^ = nxn−^1. When differentiating, results can be expressed in a number of ways. For example, (i) if y = 3x^2 then dy dx = 6x, (ii) if f (x) = 3x^2 then f ′(x) = 6x, (iii) the differential coefficient of 3x^2 is 6x. For example, if f (x) =
x = x (^12) , then f ′(x) =
x (^12) − 1 = 12 x−^ (^12) =
x
You Try It: Using the general rule, differentiate the following with respect to x : (a) f (x) = 5x^7 (b) f (x) =
x^2
- For any constant c, (cf (x))′^ = cf ′(x). For example, (5x^3 )′^ = 5(x^3 )′^ = 5(3x^2 ) = 15x^2.
- The derivative of a sum (difference) is the sum (difference) of the derivatives,
(f (x) ± g(x))′^ = f ′(x) ± g′(x).
For example,
(3x^5 − 2 x^2 + 1)′^ = (3x^5 )′^ − (2x^2 )′^ + 1′^ = 3(x^5 )′^ − 2(x^2 )′^ + 0 = 3(5x^4 ) − 2(2x) = 15x^4 − 4 x.
- Product Rule (f (x)g(x))′^ = f (x)g′(x) + g(x)f ′(x).
- Quotient Rule (^) ( f (x) g(x)
g(x)f ′(x) − f (x)g′(x) (g(x))^2
- Parametric Equations. If the coordinates (x, y) of a point P on a curve are given as functions x = f (u) and y = g(u) of a third variable or parameter u, the equations x = f (u) and y = g(u) are called parametric equations of the curve. For example, x = 12 t, y = 4 − t^2 or x = cos θ, y = 4 sin^2 θ. The First Derivative dy dx is given by
dy dx
dy/du dx/du
The Second Derivative d^2 y dx^2 is given by
d^2 y dx^2
d du
dy dx
du dx
Example: Find
dy dx and
d^2 y dx^2 given x = θ − sin θ and y = 1 − cos θ.
Solution: Note that dx dθ = 1 − cos θ and dy dθ = sin θ, so
dy dx
dy/dθ dx/dθ
sin θ 1 − cos θ
Also d^2 y dx^2
d dθ
sin θ 1 − cos θ
dθ dx = cos θ − 1 (1 − cos θ)^2
1 − cos θ
(1 − cos θ)^2
Example: Find dy dx
and d^2 y dx^2
given x = et^ cos t and y = et^ sin t.
Solution: Note that dx dt = et(cos t − sin t) and dy dt = et(sin t + cos t), so
dy dx
dy/dt dx/dt
sin t + cos t cos t − sin t
Also d^2 y dx^2
d dt
sin t + cos t cos t − sin t
dt dx =
(cos t − sin t)^2
et(cos t − sin t)
et(cos t − sin t)^3
Theorem 1.1.1. If f (x) = c is a constant function, then f ′(x) = 0 for all real numbers x.
Proof. Since f g
= f
g
, we have
( f g
(x) =
f ·
g
(x)
= f ′(x)
g(x)
- f (x)
g
(x)
f ′(x) g(x)
- f (x)
g′(x) (g(x))^2
f ′(x)g(x) − f (x)g′(x) (g(x))^2
Example: f (x) = x^2 − 1 x^2 + 1 , then f ′(x) = (x^2 + 1)(2x) − (x^2 − 1)(2x) (x^2 + 1)^2
4 x (x^2 + 1)^2
Example: If f (x) = x x^2 + 1 , then f ′(x) = (x^2 + 1) − x(2x) (x^2 + 1)^2
1 − x^2 (x^2 + 1)^2
Example: If f (x) =
x , then f ′(x) = −
x^2
Example: If x = cos t and y = t sin t, find dy dx
Solution:
dy dx
d dt (t sin t) d dt (cos t)
sin t + t cos t − sin t
1.2 Derivatives of Trigonometric Functions
Recall : lim h→ 0 sin h h = 1 and lim h→ 0 1 − cos h h
(sin x)′^ = lim h→ 0
sin(x + h) − sin x h = lim h→ 0
sin x cos h + cos x sin h − sin x h = lim h→ 0
sin x(cos h − 1) + cos x sin h h = lim h→ 0
− sin x (1 − cos h) h
cos x sin h h
= − sin x(0) + cos x(1).
Hence (sin x)′^ = cos x.
Example: Find dy dx if y = x^3 sin x.
Solution: dy dx
= (x^3 sin x)′^ = x^3 (sin x)′^ + sin x(x^3 )′^ = x^3 cos x + 3x^2 sin x.
Example: Determine whether the function,
f (x) =
x^3 sin
x
, if x 6 = 0 0 , if x = 0,
is differentiable at x = 0.
Solution: Observe that
lim h→ 0
h^3 sin (^) h^1 − 0 h = lim h→ 0 h^2 sin
h
Logarithmic Functions. Assume a > 0 and a 6 = 1. If ay^ = x, then define y = loga x. Let ln x ≡ loge x (ln x is called the natural logarithm of x).
Basic Properties of Logarithms
- loga 1 = 0 (In particular, ln 1 = 0).
- loga a = 1 (In particular, ln e = 1).
- loga uv = loga u + loga v.
- loga
(u v
= loga u − loga v.
- loga ur^ = r loga u.
Derivatives of ln x and ex^ are d dx ex^ = ex^ and d dx ln x =
x
. Also d dx (ax) = ax^ ln x, a > 0.
that, if x is the variable, then x^3 − x^2 is applied first and sin applied next. So it must be that
g(x) = x^3 − x^2 and f (s) = sin s. Notice that d ds f (s) = cos s and d dx g(x) = 3x^2 − 2 x. Then
sin(x^3 − x^2 ) = f ◦ g(x)
and
d dx (sin(x^3 − x^2 )) = d dx (f ◦ g(x))
=
[
df ds (g(x))
]
d dx g(x)
= cos(g(x)) · (3x^2 − 2 x) = [cos(x^3 − x^2 )] · (3x^2 − 2 x).
Example: Calculate the derivative d dx ln
x^2 x − 2
Solution: Let h(x) = ln
x^2 x − 2
. Then h = f ◦ g, where f (s) = ln s and g(x) = x^2 x − 2 . So d ds f (s) =
s and d dx g(x) = (x − 2) · 2 x − x^2 · 1 (x − 2)^2
x^2 − 4 x (x − 2)^2
. As a result,
d dx h(x) = d dx (f ◦ g)
=
[
df ds (g(x))
]
d dx g(x)
g(x)
x^2 − 4 x (x − 2)^2 =
x^2 x − 2
x^2 − 4 x (x − 2)^2
x − 4 x(x − 2)
You Try It: Calculate the derivative of tan(ex^ − x).
1.4 Continuity and Differentiation
What is the relationship between continuity and differentiation? It appears that functions that have derivatives must be continuous.
Theorem 1.4.1. If a function f is differentiable at a point x, then it is continuous at x.
Proof. We want to show that f is continuous at x, i.e., lim t→x f (t) = f (x) or lim h→ 0 f (x + h) = f (x),
where h = t − x. It will be sufficient to show that lim h→ 0 [f (x + h) − f (x)] = 0.
Now,
lim h→ 0 [f (x + h) − f (x)] = lim h→ 0
[
f (x + h) − f (x) h
]
h
= (^) hlim→ 0 f (x + h) − f (x) h lim h→ 0 h = f ′(x) · 0 = 0 ,
because f ′(x) is finite. Thus f is continuous at x.
Converse is false: For example, the function f (x) = |x| is continuous at x = 0, but it is not differentiable there.
1.5 Higher Order Derivatives
If f (x) is differentiable in an interval, its derivative is given by f ′(x), y′^ or dy dx where y = f (x).
If f ′(x) is also differentiable in the interval, its derivative is denoted by f ′′(x), y′′^ or d dx
dy dx
d^2 y dx^2
Similarly, the nth derivative of f (x), if it exists, is denoted by f (n), y(n)^ or dny dxn^
where n is called
the order of the derivative.
Example: Let y = f (x) = 12 x^4 − 3 x^2 + 1.
Solution: Derivative y′^ = f ′(x) = d dx (^12 x^4 − 3 x^2 + 1) = 2x^3 − 6 x.
Second derivative y′′^ = f ′′(x) = d^2 y dx^2
d dx (2x^3 − 6 x) = 6x^2 − 6.
Third derivative y′′′^ = f ′′′(x) = d^3 y dx^3
d dx
(6x^2 − 6) = 12x.
Fourth derivative y(4)^ = f (4)(x) = d^4 y dx^4
d dx (12x) = 12.
Solution: From the above example, we already know that the first derivative is
dy dx
x y
Hence by the Quotient Rule
d^2 y dx^2
d dx
x y
y · 1 − x · dy dx y^2
y − x
x y
y^2 Substituting for dy dx
= − y^2 + x^2 y^3
Noting that x^2 + y^2 = 4 permits us to write the second derivative as
d^2 y dx^2
y^3
Example: Find dy dx if sin y = y cos 2x.
Solution:
d dx
sin y = d dx
y cos 2x
cos y dy dx = y(− sin 2x · 2) + cos 2x dy dx (cos y − cos 2x) dy dx = − 2 y sin 2x dy dx
2 y sin 2x cos y − cos 2x
Example: Find y′, given xy + x − 2 y − 1 = 0.
Solution: We have
x d dx (y) + y d dx (x) + d dx (x) − 2 d dx (y) − d dx
d dx
or xy′^ + y + 1 − 2 y′^ = 0, then y′^ = 1 + y 2 − x
Example: Find y′, given x^2 y − xy^2 + x^2 + y^2 = 0.
Solution:
d dx
(x^2 y) − d dx
(xy^2 ) + d dx
(x^2 ) + d dx
(y^2 ) = 0
x^2 d dx (y) + y d dx (x^2 ) − x d dx (y^2 ) − y^2 d dx (x) + d dx (x^2 ) + d dx (y^2 ) = 0.
Hence, x^2 y′^ + 2xy − 2 xyy′^ − y^2 + 2x + 2yy′^ = 0 and y′^ = y^2 − 2 x − 2 xy x^2 + 2y − 2 xy
Example: Find y′^ and y′′, given x^2 − xy + y^2 = 3.
Solution:
d dx
(x^2 ) − d dx
(xy) = d dx
(y^2 ) = 2x − xy′^ − y + 2yy′^ = 0. So y′^ = 2 x − y x − 2 y
Then
y′′^ =
(x − 2 y) d dx (2x − y) − (2x − y) d dx (x − 2 y) (x − 2 y)^2
(x − 2 y)(2 − y′) − (2x − y)(1 − 2 y′) (x − 2 y)^2
3 xy′^ − 3 y (x − 2 y)^2
3 x
2 x − y x − 2 y
− 3 y (x − 2 y)^2
6(x^2 − xy + y^2 ) (x − 2 y)^2 =
(x − 2 y)^2
You Try It: Find y′′, given x^3 − 3 xy + y^3 = 1.
1.7 Logarithmic Differentiation
Take natural logarithm (ln) both sides, differentiate implicitly and solve for y′.
Example: Compute y′^ if y = x^2
7 x − 14 (1 + x^2 )^4
Applications of the Derivative
1.9 The Mean Value Theorem
Suppose f (x) is continuous on [a, b] and differentiable on (a, b). Then, there exists a c in (a, b) at which the tangent line is parallel to the secant line joining the points (a, f (a)) and (b, f (b)), i.e., at
which f ′(c) = f (b) − f (a) b − a
OR
If f (x) is continuous in [a, b] and differentiable in (a, b), then there exists a point c in (a, b) such that
f ′(c) = f (b) − f (a) b − a , a < c < b.
The word mean in The Mean Value Theorem refers to the mean (or average) rate of change of f in the interval [a, b].
If f (a) = f (b) = 0, then the theorem says that there exists a c in (a, b) at which f ′(c) = 0. The
graphs suggest that there must be at least one point on the graph, that corresponds to a number c in (a, b), at which the tangent is horizontal. This special case of the Mean Value Theorem is called Rolle’s Theorem 1.
Example: Consider f (x) =
x − 1 on [2, 5], f (x) is continuous when x − 1 ≥ 0, i.e., x ≥ 1. In
particular, f (x) is continuous on [2, 5] and f ′(x) =
x − 1
, so differentiable when x > 1. In
particular, f (x) is differentiable on (2, 5).
f (b) − f (a) b − a
f (5) − f (2) 5 − 2
The Mean Value Theorem asserts that, for some c in (2, 5), f ′(c) =
. Let us find it.
f ′(x) =
x − 1
x − 1 = 3 4(x − 1) = 9 x − 1 =
x =
Notice that
is in (2, 5), so we may take c =
Example: Show that if f (x) = tan x on the interval 0 ≤ x ≤ k where k < π 2 , then tan k ≥ k. (^1) Michel Rolle, a French mathematician (1652-1719)
Then, from a < c < b we have
a < c < b =⇒
b
c
a 1 b
ln b − ln a b − a
a
b − a b < ln b − ln a < b − a a =⇒ 1 − a b < ln
b a
b a
Now, ln(1.2) = ln
= ln
. Therefore a = 5 and b = 6. Substituting in
1 −
a b < ln
b a
b a − 1, we have
< ln
< ln 1. 2 <
1.10 Indeterminate Forms
Happens when lim x→a
f (x) g(x)
tends to
or
as x → a. Think of the situation lim x→a
f (x) g(x)
where f (x) and g(x) are differentiable (and therefore continuous so f (a) = lim x→a f (x) = 0 and
g(a) = lim x→a g(x) = 0.), then
xlim→a
f (x) g(x) = (^) xlim→a f (x) − f (a) g(x) − g(x)
= lim x→a
f (x) − f (a) x − a g(x) − g(a) x − a
(provided the denominator is not zero)
lim x→a
f (x) − f (a) x − a
x^ lim→a
g(x) − g(a) x − a
=
f ′(a) g′(a)
lim x→a f ′(x)
xlim→a g′(x)^
(provided f ′(x) and g′(x) are also continuous.)
Example:
x^ lim→ 2
x^2 − 4 x − 2 = lim x→ 2 (x^2 − 4)′ (x − 2)′^ = lim x→ 2 2 x 1
Theorem 1.10.1. If lim f (x) g(x) = lim f ′(x) g′(x) provided that lim f (x) g(x) is of the type
or
, this is
called L’Hˆopital’s Rule: if either lim f (x) g(x)
or
Examples: (a) lim x→ 0
sin 2x sin 5x = lim x→ 0
2 cos 2x 5 cos 5x
(b) (^) xlim→∞
e^3 x x = lim x→∞
3 e^3 x 1
(c) lim x→ 0 ex^ − 1 x^3 = lim x→ 0 ex 3 x^2
The form ∞ − ∞. A given limit that is not immediately
or
can be converted to one of these forms by combination of algebra and a little cleverness.
Example: Evaluate lim x→ 0
[
1 + 3x sin x
x
]
Solution: We note 1 + 3x sin x → ∞ and
x → ∞. However, after writing the difference as a single
fraction, we recognize the form
xlim→ 0
[
1 + 3x sin x
x
]
= (^) xlim→ 0 3 x^2 + x − sin x x sin x = (^) xlim→ 0 6 x + 1 − cos x x cos x + sin x = (^) xlim→ 0 6 + sin x −x sin x + 2 cos x =
The form 0 · ∞. By suitable manipulation, L’Hˆopital’s Rule can sometimes be applied to the limit form 0 · ∞.
Example: Evaluate lim x→∞ x sin
x
Solution: Write the given expression as
xlim→∞
sin
x
x