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1.
Calculus: Differentiation
In Physics we are often interested in how a quantity changes with time or distance, or in
response to changing some other physical quantity (e.g., temperature). Mathematically,
we can represent such a relationship using algebraic notation. For instance, suppose an
object travels a distance y in a time t and that y(t) = t2 tells us the relationship between
the two. Graphically, we can draw this on a two-dimensional plot of y versus t.
Fig. 1.1.— A relationship between distance travelled (y) and time (t). The gradient
(slope) of the straight line tangent to the curve y(t) at the point P is equal to the
instantaneous velocity (speed plus direction, which is given by the sign of the slope) of
the object at that point and time.
Suppose we want to work out how fast the object is travelling at any moment in time.
Speed (or velocity, if we care about the sign) is defined as the rate of change of distance
with respect to time—in other words, how far an object travels in a given time interval,
divided by the time. But unless the relationship between distance and time is linear (i.e.
it looks like a straight line when we plot y against t), the speed is changing all the time
and a measurement of it will be different in different time intervals. What we need to
know in general is how to find the instantaneous speed/velocity at an arbitrary instant
in time. This is given at any point by the slope, or gradient, of a straight line that just
touches (i.e., is tangent to) the graph of y(t) at that point (see Figure 1.1).
1.1.
Mathematical definition of differentiation
The process of differentiation is what tells us the instantaneous slope of the relationship
between two physical quantities. Suppose we have a linear relationship y(t) = 3t (sketched
in the left part of Figure 1.2). This relationship has the same slope (i.e., 3) at all times. If
y is measured in metres and t in seconds, then the velocity is a constant 3 m s−1 . But now
consider the relationship y(t) = t2 again (Figure 1.2, right). Here the slope or gradient of
the tangent to the curve changes with time.
The instantaneous gradient of a function at any point is known as the derivative of the
function. Here, the derivative of position, y, with respect to time, t, is our instantaneous
2
velocity. It is represented mathematically by the notation
dy
, and in general (i.e., for
dt
any other than linear y–t relationships) it will itself be a function of t. It is evaluated
by finding the slope of a straight line extending from a point y(t) on the curve at any
particular time t to a second point y(t + h) on the curve at a second time t + h, and
then letting these two points come arbitrarily close to each other by taking the limit that
h → 0 [so that t + h → t and — provided the function is continuous — y(t + h) → y(t)].
In this limit, the slope of the straight line between [t, y(t)] and [t + h, y(t + h)] approaches
the slope of the tangent to the curve at time t.
Fig. 1.2.— The plot on the left shows a linear relationship between position y and time
t, which has a constant gradient (and thus a constant velocity) equal to +3. The plot
on the right shows a curved relationship where the instantaneous gradient of the curve
changes with time, so that velocity is a function of time.
The slope of the straight line connecting any two points at times t and t + h is given by
±
£
¤±£
¤
“rise-over-run”: ∆y ∆t = y(t + h) − y(t) (t + h) − t . Taking the limit h → 0 then
gives the formal definition for the derivative of any function y(t) as
·
·
¸
¸
y(t + h) − y(t)
y(t + h) − y(t)
dy
= lim
.
= lim
h→0
h→0
dt
(t + h) − t
h
1.2.
Examples of differentiation
For the relationship y(t) = t2 between position and time:
·
¸
·
¸
· 2
¸
dy
y(t + h) − y(t)
(t + h)2 − t2
t + 2th + h2 − t2
= lim
= lim
= lim
h→0
h→0
h→0
dt
h
h
h
·
¸
2th + h2
= lim
= lim [2t + h] = 2 t .
h→0
h→0
h
Thus, the velocity of the object at time t in this case is just twice the value of the time.
More generally, irrespective of any physical interpretation of the variable t as a time, or
the variable y as a position, we have that
d 2
(t ) = 2 t .
dt
3
1
instead:
t
¸
¶¸
· µ
·
1
1
1
dy
y(t + h) − y(t)
= lim
= lim
−
h→0 h
h→0
dt
h
t+h t
·
·
¸
¸
¸
·
1
−1
1 t − (t + h)
= lim
= lim − 2
= lim
h→0 t (t + h)
h→0
h→0 h
(t + h) t
t + ht
Now, consider y(t) =
= −
1
.
t2
Since 1/t = t−1 and 1/t2 = t−2 , we can equivalently (and more conveniently, in fact) write
d ¡ −1 ¢
= −t−2 .
t
dt
1.3.
Standard derivatives
So far we have used the symbols y and t to represent the dependent and independent
variables – i.e. the value of y depends on the value of t. But the basic ideas, and the rules
and definitions that follow now, apply whatever symbols are being used and whatever
physical quantities are being considered. It is standard to use x to denote a generic
independent variable, and this is the convention that we shall follow for the most part.
Whether x is meant to represent a time t, a position y, an angle θ, a temperature T ,
etc. is a detail to be decided in the context of a specific problem.
The Keele Handbook of Mathematics, Physics and Astronomy contains listings of a large
number of standard derivatives—that is, the derivatives of many elementary and some
more complicated functions f (x), which you do not have to prove every time you encounter
them. Unless you are specifically told otherwise, you can simply look up these standard
derivatives in the Handbook to use them as necessary. But it won’t hurt to memorise
the most basic (and most common) of them; and this will happen with several of them
anyway, just through repeated use. Some of the most important standard derivatives are:
Powers of x
The derivative of any power of x is given by
d
dx
(xn) = nxn−1 .
The two examples we looked at just above, where we found dy/dt for y = t2 and y = 1/t,
are special cases of this rule for integer exponents n = 2 and n = −1. In general, though,
the power n does not have to be an integer; it can be any real number whatsoever (but it
must be a constant number, not itself a function of x or any other variable).
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Trigonometric functions (sine and cosine)
We very often need to use the derivatives of trigonometric functions, and it is
important that you become very familiar with these. At the most basic level, it is actually
enough to think about sin(x) and cos(x) only; more complicated functions—including
tan(x), products and powers of trigonometric functions, and functions of arguments more
complicated than just x—can be dealt with using the “rules” of differentiation that we
discuss in Section 1.4 below.
Consider y(x) = sin(x). Its derivative is obtained from first principles as
·
·
¸
¸
y(x + h) − y(x)
sin(x + h) − sin x
dy
= lim
= lim
h→0
h→0
dx
h
h
·
¸
¸
·
sin x cos h + cos x sin h − sin x
cos x sin h
sin x (cos h − 1)
= lim
+
= lim
h→0
h→0
h
h
h
·
¸
¸
·
sin h
cos h − 1
+ cos x × lim
(1.1)
= sin x × lim
h→0
h→0
h
h
The second line here has used the relation sin(A + B) = sin A cos B + cos A sin B (which
can be found in the Handbook of Mathematics, Physics and Astronomy). On the third
line, there are two non-trivial limits to deal with. Both of these can be evaluated by
looking at an isosceles triangle with two sides of length 1 and an angle h (measured in
radians) between these two sides, inscribed in a circle of radius 1. Let the third side
of the isosceles triangle have length H. This set-up is shown in Figure 1.3: first for a
relatively large angle h, and just below that a blow-up of the case when the angle h is
very small (i.e., |h| ≪ 1 or h → 0).
It is apparent from the upper part of Figure 1.3 that: (1) the angle h is subtended
by an arc of length h on the perimeter of the unit-radius circle (recalling the definition
of radian); and (2) the length H of the third side of the inscribed triangle is given by
(recalling Pythagoras’ theorem, and sin2 h + cos2 h = 1)
¡
¢
H 2 = (sin h)2 + (1 − cos h)2 = sin2 h + 1 − 2 cos h + cos2 h = 2 − 2 cos h .
(1.2)
This equality holds for any value of the angle h, large or small.
Getting now to the limit of very small h → 0, the lower part of Figure 1.3 makes it clear
that: (1) the vertical line segment with length sin h must be approximately equal to the
base H of the isosceles triangle; and (2) the base H must in turn be approximately equal
to the arc length h subtended by the angle h. That is:
for |h| ≪ 1 :
sin h ≈ h
=⇒
sin h
≈ 1.
h
This so-called small-angle approximation, sin h ≈ h for |h| ≪ 1 (again: in radians!),
is an extremely useful approximation in its own right. It is routinely employed in analyses
that need not have anything to do with differentiation. It (and so, of course, sin h/h ≈ 1)
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Fig. 1.3.— Diagram justifying the small-angle approximations for sine and cosine func³
h2 ´
for angles with values |h| ≪ 1 in radians.
tions: sin(h) ≈ h and cos(h) ≈ 1 −
2
is most definitely an approximate, not exact, equality; but it gets better and better as |h|
gets smaller and smaller, and ultimately in the limit that h → 0 we have exactly,
¸
·
sin h
= lim [ 1 ] = 1 .
lim
h→0
h→0
h
This is one of the limits needed in equation (1.1) to find the derivative of sin x with respect
to x. To find the other limit, which involves cos h, we make use of equation (1.2). Again,
for h → 0 we have H ≈ h, and therefore H 2 ≈ h2 , or
for |h| ≪ 1 :
2−2 cos h ≈ h2
=⇒
cos h ≈ 1−
h2
2
=⇒
cos h − 1
h
≈ − .
h
2
The small-angle approximation, cos h ≈ 1 − h2 /2 for |h| ≪ 1, is again just that—
approximate—but again it gets better as h gets smaller, and again the limit as h → 0 is
exact:
·
¸
·
¸
cos h − 1
h
lim
= lim −
= 0.
h→0
h→0
h
2
Putting both of these limits into equation (1.1), then, we have our main result for the
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derivative of the sine function:
d
dx
(sin x) = cos x .
A similar analysis, which makes use of the same limits for (cos h − 1) /h and (sin h) /h
as h → 0, further shows that the derivative of the cosine function is
d
dx
(cos x) = − sin x .
In both cases, it is always understood that the angle x is in radians.
The exponential function
The derivative of the exponential function is
d
dx
(ex) = ex ,
where e = 2.71828 . . . is Napier’s constant. Thus, ex is its own derivative. This is a
large part of what makes e such an important number in Physics (and Mathematics), and
it is essential that you know how to work with exponentials and their derivatives. Note
that another standard way of writing ex is as exp(x).
A proof that d(ex )/dx = ex is straightforward if we make use of one definition of e itself,
as the limit,
e ≡ lim (1 + h)1/h .
h→0
(This was discovered in the 17th century by Jacob Bernoulli, in his solution to a problem
about compound interest.) Therefore, for h “small enough” we can write
e ≈ (1 + h)1/h
for |h| ≪ 1 :
=⇒
eh ≈ 1 + h .
These are only approximate equalities for any value of h that is actually non-zero; but
the approximations get increasingly accurate as |h| gets smaller and are exact in the limit
that h → 0. The derivative of ex with respect to x follows directly:
· x+h
¸
· x h
¸
·
¸
h
e
− ex
e e − ex
d x
x e −1
(e ) = lim
= lim
= lim e
h→0
h→0
h→0
dx
h
h
h
eh − 1
= e lim
h→0
h
x
= ex .
·
¸
x
−→ e
·
(1 + h) − 1
h
¸
x
= e
·
h
h
¸
7
The natural logarithm
The derivative of the natural logarithm of x is
d
dx
(ln x) =
1
x
(for x > 0 only).
Recall that the natural logarithm is the inverse function of the exponential function, i.e.,
ln(ex ) = x and eln x = x . We shall use this fact below to prove that d(ln x)/dx = 1/x.
1.4.
Rules of differentiation
In the following, u ≡ u(x) and v ≡ v(x) are any two (continuous) functions of an independent variable x, and k is any constant (a real or a complex number).
Sum rule :
Factor rule :
Product rule :
Chain rule :
d
dx
(u + v) =
d
dx
d
dx
du
dx
(k u) = k
(u v) =
dy
dx
=
dv
+
(1.3)
dx
du
(1.4)
dx
µ
du
dx
dy
du
¶
×
v + u
du
dx
µ
dv
dx
¶
(1.5)
(1.6)
All of these rules can be derived from first principles by applying the basic definition of a
derivative and doing some algebra involving limits. The “sum rule” and the “factor rule”
together express the fact that differentiation is a linear operation on functions—that is,
¤
d£
dv
du
+ k2 . This is of further mathematical interest, and it
k1 u(x) + k2 v(x) = k1
dx
dx
dx
also has important consequences for the structure of physical theories such as quantum
mechanics. We will come to it again in our study of differential equations later in this
course, but for the moment it is more or less a note in passing.
The product and chain rules can be used, sometimes together, to derive other “rules” of
differentiation, which are in fact not as fundamental. One of these is the so-called quotient
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rule. Again, if u(x) and v(x) are any two continuous functions of x, then:
µ ¶
d u
d
=
(u v −1 )
dx
dx v
du
d
(using the product rule)
= u (v −1 ) + v −1
dx
dx
dv
du
d
+ v −1
(using the chain rule)
= u (v −1 )
dv
dx
dx
dv
du
= u (−v −2 )
+ v −1
dx
dx
du/dx
−u × (dv/dx)
+
=
v2
v
v × (du/dx) − u × (dv/dx)
.
=
v2
This illustrates how knowing the derivatives of just a few elementary functions (powers,
exponentials, logarithms and trigonometric functions) plus the rules in equations (1.3)
through (1.6) will allow you to differentiate any “well-behaved” function of a single variable.
Using the chain rule
We very often encounter “functions of functions” of some variable x, which require the
chain rule to find their derivatives. For example, consider y = sin(t3 ). To make this look
simpler (i.e., more like a standard derivative that we already know), we define the function
u(t) = t3 . We can then write y = sin u, and we know immediately that dy/du = cos u.
Moreover, the derivative of u = t3 with respect to t is easy to compute: du/dt = 3t2 . The
chain rule pulls everything together:
d
du
d
(sin t3 ) =
(sin u) ×
= cos u × (3t2 ) = cos t3 × (3t2 ) = 3t2 cos t3 .
dt
du
dt
As another example, let us use the chain rule to prove the standard derivative d(ln x)/dx =
1/x by starting from the standard derivative of the exponential function, d(ex )/dx = ex ,
taken as a given. To do this, we first note that, by definition, ln(ex ) = x, so differentiating
both sides of this trivial equation tells us that
d
d
[ln(ex )] =
(x) = 1 .
dx
dx
Now we use the chain rule to manipulate the left-hand side of this equality. Defining
u(x) ≡ ex , so that du/dx = ex = u, we have
d
d
d
du
d
[ln(ex )] =
[ln u] =
(ln u) ×
=
(ln u) × u .
dx
dx
du
dx
du
This must equal 1, so
u×
d
(ln u) = 1
du
=⇒
d
du
(ln u) =
1
u
.
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But this is exactly the standard derivative that we are after, only with u chosen instead of
x to represent the independent variable (it no longer matters that we said u ≡ ex earlier
in the argument).
Using the chain and product rules together
One example of the chain and product rules working together was seen above, in the
derivation of the quotient rule for differentiation. As another example, take the function
2
y(x) = xe−ax .
2
In this case, use the product rule first, with u = x and v = e−ax , to find
dy
d
d
2
2 d
2
2
= x (e−ax ) + e−ax
(x) = x (e−ax ) + e−ax .
dx
dx
dx
dx
Next, use the chain rule to find
d −ax2
dw
(e
) by setting w = −ax2 , so that
= −2ax and
dx
dx
d w
d w
dw
d −ax2
2
(e
) =
(e ) =
(e ) ×
= ew × (−2ax) = − 2ax e−ax .
dx
dx
dw
dx
Finally, then:
dy
2
2
2
= x × (−2ax e−ax ) + e−ax = (1 − 2ax2 ) e−ax .
dx
1.5.
Velocity and acceleration, and higher-order derivatives
Suppose that y(t), v(t), a(t) are the position, velocity and acceleration of an object at
time t as it moves along the y-axis. Since velocity is the rate of change of position with
respect to time, we have
dy
;
v(t) =
dt
and since acceleration is the rate of change of velocity with respect to time, we also have
a(t) =
dv
.
dt
But this means that we have arrived at the acceleration by successive differentiation of
the position. That is,
µ ¶
d dy
d2
d d
dv
=
(y) ≡
(y) .
=
a(t) =
dt
dt dt
dt dt
dt2
Thus, we say that velocity is the first derivative of position (with respect to time) and
write this in the familiar way, v(t) = dy/dt. We can then say either that acceleration is
the (first) derivative of velocity [the notation a(t) = dv/dt says precisely this] or that it is
the second derivative of position, which is what a(t) = d2 y/dt2 denotes. Be careful here:
10
d2 y/dt2 is not the same as (dy/dt)2 —the latter is the square of the velocity, which is not
(in general) the acceleration.
If we wanted to know the rate of change of acceleration with time (a quantity known as
“jerk”), we could speak of this either as the (first) derivative of a with respect to t (i.e.,
da/dt), or as the second derivative of velocity with respect to time (i.e., d2 v/dt2 ), or as
the third derivative of position with respect to time (i.e., d3 y/dt3 ).
dn f
means “differdxn
entiate f (x) with respect to x, n times in succession” and the result is called the “nth (or
nth -order) derivative” of f . In this context, n is obviously only ever a positive integer.
In general, for a function f of an independent variable x, the notation
1.6.
Maxima and minima
One of the most useful and powerful applications of differentiation is to determine maxima
and minima—in other words, to optimise some quantity by varying another.
If f (x) is a curve with peaks and troughs then the maxima and minima occur wherever the
instantaneous gradient of the curve is zero (i.e., wherever the straight line tangent to the
curve is horizontal)—see Figure 1.4. The instantaneous gradient (or slope of the tangent
line) is the derivative df /dx, so maxima and minima occur anywhere that df /dx = 0.
There can be several minima and/or maxima (collectively referred to as stationary points
or extrema); and if there are, then any one of them need not correspond to a global
maximum or minimum of the function f (x)—just a local maximum or minimum.
Fig. 1.4.— A relationship that shows two local maxima and one local minimum of a function f (x). At each of these stationary points, the instantaneous gradient (or derivative,
df /dx) of the curve is zero. At either maximum, the gradient changes from being positive
(just to the left of the maximum) to negative (just to the right of the maximum), so the
second derivative of the function, d2 f /dx2 , is negative. At the minimum, the gradient
changes from being negative (just to the left of the minimum) to positive (just to the
right of it), so the second derivative is positive.
Solving algebraically the equation df /dx = 0 will give all the values of x where maxima or
minima occur, but it will not tell whether any one stationary point is a local maximum or a
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local minimum. To distinguish between these, we need to calculate the second derivative,
d2 f /dx2 , at any value of x where df /dx = 0. If the second derivative is negative at a
stationary point, it means that the instantaneous gradient of the function changes from
positive, to zero, to negative as x goes through the stationary point from left to right,
and this identifies the point as the top of a peak—a local maximum (see Figure 1.4). If
the second derivative is positive at a stationary point, it means that the instantaneous
gradient of the function changes from negative, to zero, to positive as x goes through
the stationary point from left to right, and this identifies the point as the bottom of a
trough—a local minimum. Therefore:
A local maximum occurs at any x value where
df /dx = 0 and d2 f /dx2 < 0 .
A local minimum occurs at any x value where
df /dx = 0 and d2 f /dx2 > 0 .
There are functions for which some values of x give both df /dx = 0 and d2 f /dx2 = 0
— a few examples are f (x) = x4 , f (x) = 1 − x4 and f (x) = x3 at the point x = 0.
Any point where df /dx = 0 is always called a stationary point, no matter what value
d2 f /dx2 takes there. A stationary point that has d2 f /dx2 = 0 as well might be a local
minimum (e.g., x4 has a minimum at x = 0), or it might be a local maximum (e.g.,
1 − x4 is maximised at x = 0), or it might simply be an inflection point which is neither
a minimum nor a maximum but a sort of instantaneous “plateau” (e.g., the behaviour
of x3 at x = 0). To distinguish between these possibilities, higher than second-order
derivatives of the function need to be examined at the stationary point.
1.7.
A sample optimisation problem
(a) A projectile launched from the ground with speed v0 , at an angle θ to the horizontal,
has (in the absence of air resistance) a horizontal range given by
R(θ) =
v02 sin(2θ)
,
g
where g is the acceleration due to gravity at the surface of the Earth. For what launch
angle, θ, is the range maximised, and what is the maximum range in terms of v0 and g?
This can be solved without calculus, of course, by simply recalling that the maximum
value of the sine function is +1, and that this value is achieved for an angle of π/2 radians
(90◦ ). Thus, the launch angle θ that maximises the range R is obtained by putting
2θmax = π/2, so that θmax = π/4 radians (or 45◦ ). The maximum range is then just
±
v02 × sin(π/2) g = v02 /g.
However, realistic maximum/minimum problems are rarely as easy as this; usually they
do require the use of calculus to identify and classify any and all stationary points of more
complicated functions. To see how this works in principle, then:
The first step in this particular problem is to differentiate R with respect to θ, using the
factor rule to take the constants v02 and g outside the derivative and then applying the
12
chain rule and recalling the standard derivative of the sine function:
·
¸
v2
dR
d v02 sin(2θ)
d
v2
2v02
= 0 ×
=
sin(2θ) = 0 × [2 cos(2θ)] =
cos(2θ) .
dθ
dθ
g
g
dθ
g
g
To find any angles that make R a (local) maximum or minimum, we set dR/dθ = 0,
which requires cos(2θ) = 0 and therefore 2θ = ±π/2, ±3π/2, ±5π/2, . . . — so, finally,
θ = ±π/4, ±3π/4, ±5π/4, . . . . Physically, we discount any negative angles and any
greater than π (i.e., 180◦ ), which correspond to “launching” the projectile into the ground
(draw a sketch). This leaves the optimal launch angle to be one or both of θ = π/4 or
θ = 3π/4 (that is, 45◦ or 135◦ ). It remains to determine the sign of d2 R/dθ2 < 0 at each
of these θ-values, since d2 R/dθ2 < 0 is required along with dR/dθ = 0 if R is to be
maximised (rather than minimised).
The second derivative of R with respect to θ is (using the chain rule)
2v02 d
4v02
d2 R
=
[cos(2θ)]
=
−
sin(2θ) ,
dθ2
g dθ
g
which is negative for θ = π/4 [since then sin(2θ) = sin(π/2) = +1] and positive for
θ = 3π/4 [because then sin(2θ) = sin(3π/2) = −1]. Thus, we conclude—as expected—
that the launch angle for maximum range is θmax = π/4 (or 45◦ ), and the value of the
maximum range is
Rmax = R(θmax ) =
v2
v02
× sin(2θmax ) = 0 .
g
g
Notice that this procedure has also given us a launch angle, θ = 3π/4 (i.e., 135◦ ), that
gives dR/dθ = 0 and d2 R/dθ2 > 0 and so yields a minimum range, of Rmin = −v02 /g.
What does this mean physically?
(b) The vertical height, y, of the same projectile is given as a function of time by
y(t) = v0 sin(θ) t −
1 2
gt .
2
Calculate the maximum height reached by the projectile, in terms of the launch angle θ.
Now the launch angle θ is considered to be fixed (at some unspecified value) and the
variable quantity is the time, t. Thus, to find the maximum height for a given θ, we
differentiate y with respect to t:
dy
= v0 sin(θ) − gt .
dt
The maximum (or minimum!) occurs when dy/dt = 0 (physically: when the velocity is
instantaneously zero), which happens at time
tmin/max =
v0 sin(θ)
.
g
13
Differentiating again, we see that d2 y/dt2 = −g, which is always negative in this case (it
is, of course, the downwards acceleration due to gravity), and so the zero in velocity indeed
corresponds to a maximum, not a minimum, in height. Putting the time at maximum
height back into the original expression for y(t) we obtain
ymax
·
¸2
1
v0 sin(θ)
v0 sin(θ)
1 v02 sin2 (θ)
− g×
.
= y(tmax ) = v0 sin(θ) ×
=
g
2
g
2
g
The maximum height naturally depends on the launch angle, θ. If θ = π/4 to give the
maximum horizontal range, then the maximum height reached by the projectile when in
±
±
±
flight is ymax = (v02 2g) × sin2 (π/4) = (v02 2g) × (1/2) = v02 4g—exactly one-quarter of
the maximum range.
1.8.
Differentiation: What you need to know
You must know how to differentiate simple mathematical functions, simple products
or ratios of these functions, and compound functions using the chain rule. These will
crop up all the time in the Physics/Astrophysics course, so the happier you are with the
differentiation the more time you can spend understanding the Physics. In addition you
must know how to find the point(s) at which a mathematical function is maximised or
minimised and use this to solve “optimisation” problems such as the example in §1.7.
Another crucial skill that you must practise is to translate a problem described in words
into graphical and/or mathematical representations, and then use your mathematics and
physics knowledge to obtain quantitative solutions. In particular, you need to be aware
of situations where one physical quantity can be obtained by differentiating another (e.g.,
if you have an expression for velocity you can differentiate it to get the acceleration and
then use Newton’s second law to deduce the applied force as a function of time).
On the next page is a set of mathematical expressions of types that will be encountered
frequently throughout the course. If you can differentiate all of these then you should
have no problem with differentiation throughout the Physics/Astrophysics degree.
14
1.8.1. Functions to differentiate
Find dy/dx for each of the following:
y = x3 + 3x2 − 2x + 1
y = x1/2 + x−1/2
y =
1
1
− 2
x x
y = sin(2x) + cos(2x)
y = sin2 (2x)
y = x sin(x)
y = e2x + e−2x
2
y = e2x + e−2x
2
y = 4 exp(sin2 x)
y = ln(x) + ln(1 + x)
y = ln(2 + x2 )
y =
(1 + x)
x2
y = x e−x − (1 + x)ex
y = x (1 +
√
x )1/2
y = 2x2 ln(2x + x3 )
It won’t hurt to attempt d2 y/dx2 for all of the above as well.