# Week 4 - City University

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```4. Calculus I: Differentiation
If the tangent is unique then the gradient of the curve at A is defined to
be the gradient of the tangent to the curve at A.
4.1 The derivative of a function
Suppose we are given a curve with a point A lying on it. If the curve is
‘smooth’ at A then we can find a unique tangent to the curve at A:
a
b
c
Consider the chord AB. As B gets
closer to A, the gradient of the chord
gets closer to the gradient of the
tangent at A.
A
A
The process of finding the general gradient function for a curve is
called differentiation.
A
B
A
Here the curve in (a) is smooth at A, but the curves in (b) and (c) are
not.
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MA1601 Week 4
Anton Cox (City University)
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δy
(x,y)
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Example 4.1.1: Take f (x) = c, a constant function.
(x+δx,y+δy)
For y = f (x), the gradient function is
defined by
δy
f (x + δx) − f (x)
lim
= lim
.
δx
δx→0 δx
δx→0
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δx
At every x the gradient is 0, so f 0 (x) = 0 for all x.
Or
f (x + δx) − f (x)
c−c
=
= 0.
δx
δx
Example 4.1.2: Take f (x) = ax.
0
We denote the gradient function by dy
dx or f (x), and call it the derivative
of f . This is not the formal definition of the derivative, as we have not
explained exactly what we mean by the limit as δx → 0. But this
intuitive definition will be sufficient for the basic functions which we
consider.
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Example 4.1.3: Take f (x) = x 2 .
At every x the gradient is a, so f 0 (x) = a for all x.
Or
f (x + δx) − f (x)
a(x + δx) − ax
aδx
=
=
= a.
δx
δx
δx
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δx)2
−
δx
x 2 + 2xδx + (δx)2 − x 2
=
δx
δx(2x + δx)
=
= 2x + δx.
δx
=
(x +
=
The limit as δx tends to 0 is − x12 , so f 0 (x) = − x12 .
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Example 4.1.5: Take f (x) = x n with n ∈ N and n &gt; 1.
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Example 4.1.6: f (x) = sin x.
Recall that
n
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1
1
1
−
δx x + δx
x
x − (x + δx)
=
(δx)(x + δx)x
−δx
−1
=
=
.
(δx)(x + δx)x
(x + δx)x
f (x + δx) − f (x)
δx
x2
The limit as δx tends to 0 is 2x, so f 0 (x) = 2x.
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Example 4.1.4: Take f (x) = x1 .
Now we need to consider the second formulation, as we cannot simply
read the gradient off from the graph.
f (x + δx) − f (x)
δx
MA1601 Week 4
We use the identity for sin A + sin B.
n
a − b = (a − b)(a
n−1
+a
n−2
b+a
n−3 2
b + &middot;&middot;&middot; + b
n−1
)
f (x + δx) − f (x) = 2 sin
and so
an − b n
= an−1 + an−2 b + an−3 b2 + &middot; &middot; &middot; + bn−1
a−b
where the sum has n terms. As a → b we have
n
a − bn
= lim (an−1 + an−2 b + an−3 b2 + &middot; &middot; &middot; + bn−1 ) = nbn−1 .
lim
a−b
a→b
a→b
and so
θ→0
and so
f 0 (x) = lim
Hence f 0 (x) = nx n−1 .
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δx
cos x +
2
We need the following fact (which we will not prove here):
δx→0
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sin δx
f (x + δx) − f (x)
δx
2
=
cos x +
.
δx
δx
2
2
lim
If a = x + δx and b = x then
n
f (x + δx) − f (x)
a − bn
lim
= lim
= nbn−1 = nx n−1 .
δx
a−b
δx→0
a→b
δx
2
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sin
δx
2
δx
2
sin θ
=1
θ
δx
cos x +
= cos(x).
2
MA1601 Week 4
4.2 Differentiation of compound functions
Some standard derivatives, which must be memorised:
Once we know a few basic derivatives, we can determine many others
using the following rules:
Let u(x) and v (x) be functions of x, and a and b be constants.
f 0 (x)
——
kx k −1
ex
f (x)
——
xk
ex
ln x
sin x
cos x
tan x
cosec x
sec x
cot x
Function
————
au &plusmn; bv
uv
1
x
cos x
− sin x
sec2 x
− cosec x cot x
sec x tan x
− cosec2 x
Sum and difference
Product
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Anton Cox (City University)
v du
−u dv
dx
dx
v2
du dz
.
dz dx
u
v
Quotient
Composite
Some of these results can be derived from the results in the following
sections, or from first principles. However it is much more efficient to
know them.
Derivative
—————
dv
a du
dx &plusmn; b dx
dv
v du
+
u
dx
dx
u(v (x))
where z = v (x).
The final rule above is known as the chain rule and has the following
special case
u(ax + b) a du
dx (ax + b)
For example, the derivative of sin(ax + b) is a cos(ax + b).
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Example 4.2.1: Differentiate
y = 2x 5 − 3x 3 +
Example 4.2.3: Differentiate
4
.
x2
y = x 2 ln(x + 3).
dy
8
= 10x 4 − 9x 2 − 3 .
dx
x
dy
x2
= 2x ln(x + 3) +
.
dx
x +3
Example 4.2.2: Differentiate
Example 4.2.4: Differentiate y = e5x .
x2 − 1
y= 2
.
x +1
Set z = 5x, then
dy
dy dz
=
= ez 5 = 5e5x .
dx
dz dx
dy
(x 2 + 1)2x − (x 2 − 1)2x
4x
=
= 2
.
dx
(x 2 + 1)2
(x + 1)2
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Anton Cox (City University)
Example 4.2.7: y = cosec x =
Example 4.2.5: Differentiate y = 4 sin(2x + 3).
dy
dy dz
=
= 4 cos(z)2 = 8 cos(2x + 3).
dx
dz dx
Example√4.2.8: y = ln(x +
u = x + x 2 + 1.
As we have already noted, some of the standard derivatives can be
deduced from the others.
x 2 + 1), i.e. y = ln u where
du
(x 2 + 1)− 2
=1+
.2x
dx
2
and
so
sin x
.
cos x
dy
cos x cos x − sin x(− sin x)
cos2 x + sin2 x
1
=
=
=
= sec2 x.
dx
cos2 x
cos2 x
cos2 x
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√
1
dy
1 du
=
dx
u dx
Example 4.2.6: Differentiate
Anton Cox (City University)
1
sin x .
dy
sin x.(0) − 1. cos x
− cos x
=
=
= − cosec x cot x.
dx
sin2 x
sin2 x
Set z = 2x + 3, then
y = tan x =
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dy
1
√
=
dx
x + x2 + 1
1+ √
=
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x
x2
+1
1
√
x + x2 + 1
√
x2 + 1 + x
√
x2 + 1
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!
=√
1
x2 + 1
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Example 4.3.1: y = ln(1 + x 2 ).
Let z =
Example 4.2.9: y = x x .
dy
dx
=
2x
.
1+x 2
d2 y
dz
(1 + x 2 ).2 − 2x(2x)
2(1 − x 2 )
=
=
=
.
dx
dx 2
(1 + x 2 )2
(1 + x 2 )2
We have y = (eln x )x = e(x ln x) , i.e. y = eu where u = x ln x.
du
dy
= eu
= ex ln x (ln(x) + 1) = x x (ln(x) + 1).
dx
dx
Example 4.3.2: Show that y = e−x sin(2x) satisfies
4.3 Higher derivatives
d2 y
dy
+2
+ 5y = 0.
dx
dx 2
The derivative dy
dx is itself a function, so we can consider its derivative.
If y = f (x) then we denote the second derivative, i.e. the derivative of
dy
d2 y
00
dx with respect to x, by dx 2 or f (x). We can also calculate the higher
dy
= −e−x sin 2x + 2e−x cos 2x = e−x (2 cos 2x − sin 2x)
dx
derivatives
dn y
dx n
or f (n) (x).
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d2 y
= −e−x (2 cos 2x − sin 2x) + e−x (−4 sin 2x − 2 cos 2x)
dx 2
= e−x (−3 sin 2x − 4 cos 2x).
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Writing s for sin 2x and c for cos 2x we have
Now
y 00 + 2y 0 + 5y = e−x (−3s − 4c − 2s + 4c + 5s) = 0.
4
9
dy
=
−
dx
(1 + x)3 (1 + 3x)2
d2 y
−12
54
=
+
dx 2
(1 + x)4 (1 + 3x)3
Example 4.3.3: Evaluate
d3
dx 3
3x 2
1+
(1 + x)2 (1 + 3x)
54 &times; 9
d3 y
48
−
=
(1 + x)5 (1 + 3x)4
dx 3
at x = 0.
and substituting x = 0 we obtain that
We could use the quotient rule, but this will get complicated. Instead
we use partial fractions.
y=
d3 y
(0) = 48 − 486 = −438.
dx 3
1 + 3x 2
A
B
C
=
+
+
.
1+x
(1 + x)2 (1 + 3x)
(1 + x)2 1 + 3x
We obtain (check!) A = 0, B = −2, and C = 3.
MA1601 Week 4
Anton Cox (City University)
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Generally it is hard to give a simple formula for the nth derivative of a
function. However, in some cases it is possible. The following can be
proved by induction.
and
y 0 = a cos(ax)
y 00 = −a2 sin(ax)
y 000 = −a3 cos(ax)
y (iv ) = a4 sin(ax)
d2 y
= a2 eax .
dx 2
We can show that
We can show that
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= a sin(ax + π2 )
= a2 sin(ax + π)
= a3 sin(ax + 3π
2 )
= a4 sin(ax + 2π).
dn y
nπ
= an sin(ax +
).
dx n
2
dn y
= an eax .
dx n
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Example 4.3.5: y = sin(ax).
Example 4.3.4: y = eax .
dy
= aeax
dx
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MA1601 Week 4
Sometimes it is useful to use the Leibnitz rule, which gives a formula
for the nth derivative of a product of functions.
4.4 Differentiating implicit functions
dn
dn
n−2
n−1
d
d2
n d
n d
(f ) (g)+
(f )
(g)+&middot; &middot; &middot;
(fg) = n (f )g +
n
n−1
dx
dx
1 dx
dx
2 dx n−2 dx 2
n−1
n
dn
d
d
&middot;&middot;&middot; +
(f ) n−1 (g) + f n (g).
dx
n − 1 dx
dx
Sometimes we cannot rearrange a function into the form y = f (x), or
we may wish to consider the original form anyway (for example,
because it is simpler). However, we may still wish to differentiate with
respect to x.
Given a function g(y ) we have from the chain rule
So for example
d
d
dy
(g(y )) =
(g(y )) .
dx
dy
dx
d3
d3
d2
d
d
d2
d3
(fg) = 3 (f )g + 3 2 (f ) (g) + 3 (f ) 2 (g) + f 3 (g).
dx
dx
dx 3
dx
dx
dx
dx
As can be seen, this is very similar to the binomial theorem, and can
also be proved by induction (using the product rule).
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Example 4.4.1: x 2 + 3xy 2 − y 4 = 2.
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Example 4.4.2:
d
d 2
(x + 3xy 2 − y 4 ) =
(2) = 0.
dx
dx
2
x2
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+
3
y2
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= 12 .
Therefore we have
d 4
d
(3xy 2 ) −
(y ) = 0
dx
dx
d
dy
2x + 3y 2 + 3x (y 2 ) − 4y 3
=0
dx
dx
dy
2
3 dy
2x + 3y + 6xy
− 4y
= 0.
dx
dx
2x +
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d 2
3
d 1
( + )=
( ) = 0.
dx x 2 y 2
dx 2
Therefore we have
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d
4
3
+
=0
x 3 dx y 2
4
6 dy
− 3− 3
= 0.
x
y dx
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4.5 Differentiating parametric equations
To differentiate a parametric equation in the variable t we use
Sometimes there is no easy way to express the relationship between x
and y directly in a single equation. In such cases it may be possible to
express the relationship between them by writing each in terms of a
third variable. We call such equations parametric equations as both x
and y depend on a common parameter.
Example 4.5.1: x = t 3
dy dt
dy
=
dx
dt dx
dy
= 2t − 4
dt
Although we can write this in the form
and so
1
y = x 3 − 4x 3 + 2
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dx
= 3t 2
dt
dy
2t − 4
=
.
dx
3t 2
the parametric version is easier to work with.
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dt
1
= dx .
dx
dt
Example 4.5.1: (Continued.)
y = t 2 − 4t + 2.
2
and
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Example 4.5.2: Find the second derivative with respect to x of
x = sin θ
We have
y = cos 2θ.
dx
= cos θ
dθ
Therefore
Note: The rules so far may suggest that derivatives can be treated just
like fractions. However
d2 y
d2 y d2 t
6= 2 2
dx 2
dt dx
in general. Moreover
2 −1
d2 y
d x
6=
.
2
dx
dy 2
dy
= −2 sin 2θ.
dθ
dy
−2 sin 2θ
=
= −4 sin θ.
dx
cos θ
Now
d2 y
d
=
dx
dx 2
dy
dx
=
d
d
dθ
−4 cos θ
(−4 sin θ) =
(−4 sin θ)
=
= −4.
dx
dθ
dx
cos θ
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4.6 Tangents and normals to curves
We have already defined the value of the derivative f 0 of a function f at
a point x0 to be the gradient of f at x0 . Thus we can easily use the
derivative to write down the equation of the tangent to that point. Using
the equation for a line passing through (x0 , f (x0 )) we have that the
tangent to f at x0 is
dy
y − f (x0 ) =
(x0 )(x − x0 ).
dx
−1
dy
dx (x0 )
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Example 4.6.1: Find the equation of the tangent and normal to the
curve
y = x 2 − 6x + 5
at the point (2, −3).
We have
and hence
dy
= 2x − 6
dx
dy
dx (2)
= 4 − 6 = −2. Hence the equation of the tangent is
y + 3 = −2(x − 2) i.e. y = −2x + 1.
The normal to f at x0 is the line passing through (x0 , f (x0 ))
perpendicular to the tangent. This has equation
y − f (x0 ) =
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1
The gradient of the normal is −1
−2 = 2 , and hence the equation of the
normal is
x
1
y + 3 = (x − 2) i.e. y = − 4.
2
2
(x − x0 )
(when this makes sense).
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4.7 Stationary points and points of inflexion
If f 0 (x) &gt; 0 for a &lt; x &lt; b then f is increasing on a &lt; x &lt; b
We can tell a lot about a function from its derivatives.
Example 4.7.1:
e.g. arcs PQ, SU, UV.
Q
V
If f 0 (x) &lt; 0 for a &lt; x &lt; b then f is decreasing on a &lt; x &lt; b
U
P
R
e.g. arc QS.
T
S
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A point of inflexion is one where f 00 (x0 ) = 0 and f 00 changes sign at x0 .
A stationary point on a curve y = f (x) is a point (x0 , f (x0 )) such that
f 0 (x0 ) = 0. These come in various forms:
Type
e.g. R, T, U.
Test
f 0 (x)
f 00 (x)
Changes from + to −
Changes from − to +
No sign change
Local maximum
Local minimum
Point of inflexion
If f 00 (x) &gt; 0 for a &lt; x &lt; b then f is concave up on a &lt; x &lt; b
−ve
+ve
(see below)
e.g. arc RST.
If f 00 (x) &lt; 0 for a &lt; x &lt; b then f is concave down on a &lt; x &lt; b
e.g. Q is a max, S is a min, U is a point of inflexion.
e.g. arc PQR.
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Note that the maxima and minima above are only local. This means
that in a small region about the given point they are extremal values,
but perhaps not over the whole curve. Extremal values for the whole
curve are called global maxima or minima.
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y = 3x 4 + 8x 3 − 6x 2 − 24x + 2.
We have
Y
C
dy
= 12x 3 + 24x 2 − 12x − 24
dx
and
D
B
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Example 4.7.3: Find the stationary values and points of inflexion of
Example 4.7.2: Consider the function f on the domain X ≤ x ≤ Y
given by the graph
A
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Stationary points when
X
d2 y
= 36x 2 + 48x − 12.
dx 2
dy
dx
= 0, i.e. (check) x = 1, −1, −2.
Both A and C are local maxima, and B and D are local minima.
However the global maximum is at Y and the global minimum at X.
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y0
y 00
−0+
72
Min
+0−
−24
Max
−0+
36
Min
√
1
Points of inflexion at x = 3 (−2 &plusmn; 7), i.e. (x, y ) ≈ (0.22, −3.36) and
(x, y ) ≈ (−1.55, 12.32).
(1, −17)
(−1, 15)
(−2, 10)
−2
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−1
1
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