MTH4100 Calculus I
Lecture notes for Week 6
Thomas’ Calculus, Sections 3.5 to 4.1 except 3.7
Rainer Klages
School of Mathematical Sciences
Queen Mary University of London
Autumn 2009
Derivatives of trigonometric functions
(1) Differentiate f (x) = sin x:
• Start with the definition of f ′ (x):
sin(x + h) − sin x
f ′ (x) = lim
h→0 h
• Use sin(x + h) = sin x cos h + cos x sin h:
sin x(cos h − 1) + cos x sin h
f ′ (x) = lim
h→0 h
• Collect terms and apply limit laws:
cos h − 1 sin h
f ′ (x) = sin x lim + cos x lim
h→0 h h→0 h
cos h − 1 sin h
• Use lim = 0 and lim = 1 to conclude f ′ (x) = cos x.
h→0 h h→0 h
d
(2) A very similar derivation gives cos x = − sin x.
dx
(3) We still need d d sin x
tan x =
dx dx cos x
d d
dx
(sin x) cos x − sin x dx (cos x)
(quotient rule) =
cos2 x
cos x cos x − sin x(− sin x)
=
cos2 x
cos x + sin2 x
2
1
= =
cos2 x cos2 x
Summary: Derivatives of trigonometric functions
d
sin x = cos x
dx
d
cos x = − sin x
dx
d 1
tan x = = sec2 x
dx cos2x
d d 1
sec x = = sec x tan x
dx dx cos x
d d cos x
cot x = = − csc2 x
dx dx sin x
d d 1
csc x = = − csc x cot x
dx dx sin x
3
Derivative of composites
example: relating derivatives
y = 32 x is the same as y = 21 u and u = 3x. By differentiating
dy 3 dy 1 du
= , = , =3,
dx 2 du 2 dx
we find that
dy dy du
= .
dx du dx
Coincidence or general formula: Do rates of change multiply?
The chain rule:
examples:
(1) Differentiate x(t) = cos(t + 1).
Here: Choose x = cos u and u = t + 1 and differentiate,
dx du
= − sin u and =1.
du dt
Then
dx
= (− sin u) · 1 = − sin(t + 1) .
dt
(2)
d
sin(x2 + x) = cos(x2 + x)(2x + 1)
dx
4
Parametric equations
example:
Describe a point moving in the xy-plane as a function of a parameter t (“time”) by two
functions
x = f (t) , y = g(t) .
This may be the graph of a function, but it need not be.
The variable t is a parameter for the curve. If t ∈ [a, b], which is called a parameter
interval, then (f (a), g(a)) is the initial point, and (f (b), g(b)) is the terminal point.
Equations and interval constitute a parametrisation of the curve.
examples:
√
(1) Given is the parametrisation x = t , y = t , t ≥ 0. What is the path defined by these
equations?
Solve for y = f (x): y = t , x2 = t ⇒ y = x2 . Note that the domain of f is only [0, ∞)!
5
(2) Find a parametrisation for the line segment from (−2, 1) to (3, 5).
• Start at (−2, 1) for t = 0 by making the ansatz (“educated guess”)
x = −2 + at , y = 1 + bt .
• Implement the terminal point at (3, 5) for t = 1:
3 = −2 + a , 5=1+b.
• We conclude that a = 5 , b = 4.
• Therefore, the solution based on our ansatz is:
x = −2 + 5t , y = 1 + 4t , 0 ≤ t ≤ 1 ,
which indeed defines a straight line (why?).
A parametrised curve x = f (t), y = g(t) is differentiable at t if f and g are differentiable
at t. At a point where y is a differentiable function of x, say y = y(x), it is y = y(x(t)) and
by the chain rule
dy dy dx
= .
dt dx dt
Solving for dy/dx yields the
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example: Describe the motion of a particle whose position P (x, y) at time t is given by
x = a cos t , y = b sin t , 0 ≤ t ≤ 2π
and compute the slope at P .
• Find the equation in (x, y) by eliminating t:
Using cos t = x/a, sin t = y/b and cos2 t + sin2 t = 1 we obtain
x2 y 2
+ 2 =1,
a2 b
which is the equation of an ellipse.
dx dy
• With dt
= −a sin t and dt
= b cos t the parametric formula yields
dy dy/dt b cos t
= = .
dx dx/dt −a sin t
dy b2 x
Eliminating t again we obtain =− 2 .
dx a y
Implicit differentiation
problem: We want to compute y ′ but do not have an explicit relation y = f (x) available.
Rather, we have an implicit relation
F (x, y) = 0
between x and y.
example:
F (x, y) = x2 + y 2 − 1 = 0 .
solutions:
1. Use parametrisation, for example, x = cos t, y = sin t for the unit circle.
2. If no obvious parametrisation of F (x, y) = 0 is possible: use implicit differentiation.
example: Given y 2 = x, compute y ′.
New method by differentiating implicitly:
• Differentiating both sides of the equation gives 2yy ′ = 1.
1
• Solving for y ′ we get y ′ = 2y
.
Compare with differentiating explicitly:
√ √
• For y 2 = x we have the two explicit solutions |y| = x ⇒ y1,2 = ± x with derivatives
y1,2
′
= ± 2√1 x .
7
√
• Compare with solution above: substituting y = y1,2 = ± x therein reproduces the
explicit result.
x2 y 2
example: Find dy/dx for the ellipse, + 2 = 1.
a2 b
2x 2yy ′
1. + 2 =0
a2 b
2yy ′ 2x
2. =− 2
b2 a
b2 x
3. y ′ = − , as obtained via parametrisation in the previous lecture.
a2 y
p
application: Motivate the power rule for rational powers by differentiating y = x q using
implicit differentiation:
• write y q = xp
• differentiate: qy q−1y ′ = pxp−1
• solve for y ′ as a function of x:
p
p xp−1 p xp y py p xq p p
′
y = q−1
= q
= = = x q −1
qy qy x qx q x q
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note: Above we have silently assumed that y ′ exists! Therefore we have ‘motivated’ but
not (yet) proved this theorem!
Linearisation
“Close to” the point (a, f (a)), the tangent L(x) = f (a) + f ′ (a)(x − a) (point-slope form) is
a “good” approximation for y = f (x).
√
example: Compute the linearisation for f (x) = 1 + x at x = a = 0.
We have f (0) = 1 and with f ′ (x) = 21 (1 + x)−1/2 we get f ′ (0) = 21 , so
1
L(x) = 1 + x .
2
9
How accurate is this approximation? Magnify region around x = 0:
Why are linearisations useful? Simplify problems, solve equations analytically, . . . many
applications!
Make phrases like “close to a point (a, f (a)) the linearisation is a good approximation”
mathematically precise in terms of differentials:
L(x) = f (a) + f ′ (a)(x − a)
L(x) − f (a) = f ′ (a) (x − a)
| {z } | {z }
dy dx
Choose x = a + dx, a = x:
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Reading Assignment: read
Thomas’ Calculus, p.225-228 about Differentials
Extreme values of functions
These values are also called absolute extrema, or global extrema.
example:
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Domain abs. max. abs. min.
(a) (−∞, ∞) none 0, at 0
(b) [0, 2] 4, at 2 0, at 0
(c) (0, 2] 4, at 2 none
(d) (0, 2) none none
The existence of a global maximum and minimum is ensured by
examples:
Classify maxima and minima:
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. . . and the extension of this definition to endpoints via half-open intervals at endpoints.
note: Absolute extrema are automatically local extrema!
