1
Calculus
1.1
Gradients and the Derivative
𝑄
𝑓 (𝑥+ℎ)
𝛿𝑦 ((((𝑇
((
(
(
(
((((
𝑃
((
𝑅 𝑓 (𝑥)
𝛿𝑥
0
𝑥
𝑥+ℎ
Let 𝑃 (𝑥, 𝑓 (𝑥)) and 𝑄(𝑥+ℎ, 𝑓 (𝑥+ℎ)) denote two points on the curve of the function 𝑦 = 𝑓 (𝑥)
and let 𝑅 denote the point of intersection of the horizontal through 𝑃 with the vertical through
𝑄. Let 𝛿𝑥 = (𝑥 + ℎ) − 𝑥 = ℎ and 𝛿𝑦 = 𝑓 (𝑥 + ℎ) − 𝑓 (𝑥). Then the gradient of the chord 𝑃 𝑄 is
given by
𝛿𝑦
𝑓 (𝑥 + ℎ) − 𝑓 (𝑥)
𝑄𝑅
=
=
.
𝑃𝑅
𝛿𝑥
ℎ
Let 𝑃 𝑇 denote the tangent to the curve at 𝑃 . The gradient to the curve 𝑦 = 𝑓 (𝑥) at 𝑃 may be
approximated by the chord 𝑃 𝑄 by taking the point 𝑄 to be sufficiently close to 𝑃 . Let 𝑄 take
a succession of values taking it closer and closer to 𝑃 then the straight line through 𝑃 and 𝑄
approaches closer and closer to the tangent 𝑃 𝑇 . As 𝑄 becomes arbitrarily close to, but never
coinciding with 𝑃 , the line through 𝑃 𝑄 becomes arbitrarily close to the tangent 𝑃 𝑇 . We shall
write 𝑄 → 𝑃 , and say that ”𝑄 tends to 𝑃 ”. We shall also say that 𝑄 tends to the limit 𝑃 .
Finally, we shall say that in the limit as 𝑄 tends to 𝑃 the gradient of the chord 𝑃 𝑄 tends to the
gradient of the tangent 𝑃 𝑇 . The notion of limit may then be extended from points to numbers
by saying that in the limit as 𝑄 → 𝑃 we have that 𝛿𝑥 → 0, ℎ → 0 and 𝑓 (𝑥 + ℎ) → 𝑓 (𝑥). It is
clear now why we stipulate that 𝑃 must not coincide with 𝑄, for in this case, 𝛿𝑥 = 0, ℎ = 0,
𝑓 (𝑥 + ℎ) = 𝑓 (𝑥) and the ratio 𝑄𝑅
is indeterminate, being of the form ”0/0”, sometimes called
𝑃𝑅
an indeterminate form. We now abbreviate the phrase ”in the limit as ℎ tends to zero” by
limℎ→0 and similarly for 𝛿𝑥 tending to zero, we write lim𝛿𝑥→0 . We define the gradient of the
tangent of the curve at the point 𝑃 to be the limiting value of the gradient of the chord 𝑃 𝑄 as
𝑄 → 𝑃 and we call this the derivative of the function 𝑦 = 𝑓 (𝑥) at 𝑥. Formally,
Definition 1.1
The derivative of the function 𝑦 = 𝑓 (𝑥) at 𝑥 is
(
)
𝑓 (𝑥 + ℎ) − 𝑓 (𝑥)
′
𝑓 (𝑥) = lim
ℎ→0
ℎ
1
or equivalently,
𝑑𝑦
𝛿𝑦
= lim
𝑑𝑥 𝛿𝑥→0 𝛿𝑥
Notes
1.
if 𝑄 approaches 𝑃 from the right (left) then ℎ > 0 (ℎ < 0).
2.
the notation
1.2
𝑑𝑦
𝑑𝑥
does not indicate a ratio, it represents the limit of a ratio.
Limits
In the above diagram, 𝑦 = 𝑓 (𝑥), 𝑃 is a fixed point, 𝑄 is a variable point, so 𝑥 is a fixed number,
and ℎ is a variable. Let 𝑄 → 𝑃 from either the left (ℎ < 0) or right (ℎ > 0), and consider the
values of 𝑓 (𝑥 + ℎ) as ℎ → 0. If ℎ → 0 and ℎ > 0 (ℎ < 0) write ℎ → 0+ (ℎ → 0−). Consider the
limits 𝑙𝑖𝑚ℎ→0− 𝑓 (𝑥 + ℎ), 𝑙𝑖𝑚ℎ→0+ 𝑓 (𝑥 + ℎ). If a limit gives a definite value, the limit is said to
exist. If it does not give a definite value, the limit does not exist. For the function whose graph
is shown in the diagram, the three numbers 𝑓 (𝑥), 𝑙𝑖𝑚ℎ→0+ 𝑓 (𝑥 + ℎ), 𝑙𝑖𝑚ℎ→0− 𝑓 (𝑥 + ℎ) all exist
and are equal. In this case 𝑦 = 𝑓 (𝑥) is said to be continuous at 𝑥. If it happens that the three
numbers are not all equal or some of them do not exist, then 𝑦 = 𝑓 (𝑥) is discontinuous at 𝑥.
Many of the functions that we are familiar with are continuous, for example all polynomials,
sin𝑥, cos𝑥, exp𝑥. But, many, while being continuous for most values of 𝑥 also have some points
of discontinuity, for example rational functions (at points where the denominator is zero but
the numerator is not), tan𝑥, cosec𝑥, sec𝑥, cot𝑥.
Let 𝑓 (𝑥) = 𝑥2 and 𝑥 = 2 then 𝑙𝑖𝑚ℎ→0− (2 + ℎ)2 = 22 = 𝑙𝑖𝑚ℎ→0+ (2 + ℎ)2 . In
this case, both limits and the value of the function exist and all three numbers are equal.
Example 1.2
1.
2.
Let 𝑓 (𝑥) = 𝑥1 and 𝑥 = 0 then 𝑓 (0) is not defined. Also, if 𝑥 = ℎ and ℎ > 0 then as
ℎ → 0, ℎ1 takes larger and larger values without bound. We write 𝑓 (𝑥) → ∞ as 𝑥 → 0+.
Similarly, if ℎ < 0, ℎ1 takes negative values with larger and larger magnitude. We write
𝑓 (𝑥) → −∞ as 𝑥 → 0−. In this case, the value of the function is not defined and either
the limits do not exist, or if one allows the possibility of ±∞ as a limit, then the limits
are not equal.
3.
The function
𝑥2 + 𝑥
𝑥
is not defined at 𝑥 = 0 since division by zero is not allowed. It is not the same function
𝑓 (𝑥) =
as 𝑔(𝑥) = 𝑥 + 1. But for any value of 𝑥 other than 0, we do have 𝑓 (𝑥) = 𝑔(𝑥) = 𝑥 + 1 and
hence, since g(x) is continuous,
lim 𝑓 (𝑥) = 1
𝑥→0
i.e. as 𝑥 becomes closer and closer to 0, 𝑓 (𝑥) becomes closer and closer to 1. In this case
the value of the function does not exist, while the limits exist and are equal.
2
Now we change notation and allow 𝑥 to be a variable.
Theorem 1.3
Properties of Limits
Let 𝑐 be a fixed number, and 𝑓 (𝑥) and 𝑔(𝑥) functions of 𝑥. Then provided the limits exist
(
) (
)
lim (𝑓 (𝑥) + 𝑔(𝑥)) = lim 𝑓 (𝑥) + lim 𝑔(𝑥)
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
i.e. the limit of a sum equals the sum of the limits.
(
)(
)
lim (𝑓 (𝑥)𝑔(𝑥)) = lim 𝑓 (𝑥) lim 𝑔(𝑥)
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
i.e. the limit of a product equals the product of the limits.
The proof is beyond the scope of this course.
An important example of a limit is the following:
Theorem 1.4
For 𝜃 measured in radians, as 𝜃 → 0, we have
sin 𝜃
→ 1.
𝜃
Alternatively,
sin 𝜃
= 1.
𝜃→0
𝜃
lim
Proof: Once again, we cannot substitute 𝜃 = 0 as this gives a division by zero. We use a
geometric argument. Let the circle be of radius 𝑟.
Z
Z
Z
Z
Z
𝐷
Z
AZ
AZ
Z
Z
A
𝜃
Z 𝐶
A
Z
A
Z
𝐴
𝐵
Z
Z
Z
Z
Since the line 𝐶𝐷 is a tangent to the circle, angle 𝐴𝐷𝐶 is a right angle. Now consider the
three areas: △𝐴𝐷𝐵 < sector 𝐴𝐷𝐵 < △𝐴𝐷𝐶. The area of △𝐴𝐷𝐵 is 21 𝑟2 sin 𝜃, the area of the
𝜃
1
sector is
× 𝜋𝑟2 = 𝑟2 𝜃 and the area of △𝐴𝐷𝐶 is 12 𝑟2 tan 𝜃. Hence,
2𝜋
2
1 2
1
1
𝑟 sin 𝜃 < 𝑟2 𝜃 < 𝑟2 tan 𝜃
2
2
2
3
dividing by 12 𝑟2 gives
sin 𝜃 < 𝜃 < tan 𝜃
The first inequality gives
sin 𝜃
<1
𝜃
while the second gives
cos 𝜃 <
Now let 𝜃 → 0 then cos 𝜃 → 1 and so
Hence the result.
1.3
sin 𝜃
.
𝜃
sin 𝜃
is trapped between 1 and a number tending to 1.
𝜃
Differentiation
The process of finding the derivative of a function is called differentiation. Having defined the
derivative as the gradient of a curve, we now show how to use the definition to find derivatives
algebraically.
Example 1.5
Let 𝑦 = 𝑥2 . Then we have
(
)
)
(
𝑑𝑦
𝑓 (𝑥 + ℎ) − 𝑓 (𝑥)
(𝑥 + ℎ)2 − 𝑥2
= lim
= lim
ℎ→0
𝑑𝑥 ℎ→0
ℎ
ℎ
( 2
)
(
)
𝑥 + 2𝑥ℎ + ℎ2 − 𝑥2
2𝑥ℎ + ℎ2
= lim
= lim
= lim (2𝑥 + ℎ) = 2𝑥
ℎ→0
ℎ→0
ℎ→0
ℎ
ℎ
So, the derivative of 𝑥2 is 2𝑥.
The next example generalises this result to higher powers of 𝑥.
Example 1.6
Let 𝑦 = 𝑥𝑛 , with 𝑛 a positive integer. Then
(
)
(
)
𝑑𝑦
𝑓 (𝑥 + ℎ) − 𝑓 (𝑥)
(𝑥 + ℎ)𝑛 − 𝑥𝑛
= lim
= lim
ℎ→0
𝑑𝑥 ℎ→0
ℎ
ℎ
Expand (𝑥 + ℎ)𝑛 using the binomial theorem — only the first few terms will be needed:
( 𝑛
)
𝑥 + 𝑛𝑥𝑛−1 ℎ + 21 𝑛(𝑛 − 1)𝑥𝑛−2 ℎ2 + 16 𝑛(𝑛 − 1)(𝑛 − 2)𝑥𝑛−3 ℎ3 + ... + ℎ𝑛 − 𝑥𝑛
= lim
ℎ→0
ℎ
(
)
1
1
𝑛−1
𝑛−2
𝑛−3 2
𝑛−1
= lim 𝑛𝑥
+ 𝑛(𝑛 − 1)𝑥 ℎ + 𝑛(𝑛 − 1)(𝑛 − 2)𝑥 ℎ + ... + ℎ
ℎ→0
2
6
cancelling first the 𝑥𝑛 ’s and then an ℎ from the numerator and denominator. Now, let ℎ → 0
then all but the first term becomes zero, so
𝑑𝑦
= 𝑛𝑥𝑛−1
𝑑𝑥
Note: Cancelling the ℎ’s, eliminated the problem of division by 0 and effectively enabled us to
put ℎ = 0 in the remaining formula.
4
Example 1.7
Let 𝑦 = sin 𝑥. Then
(
)
(
)
𝑑𝑦
𝑓 (𝑥 + ℎ) − 𝑓 (𝑥)
sin (𝑥 + ℎ) − sin 𝑥
= lim
= lim
ℎ→0
𝑑𝑥 ℎ→0
ℎ
ℎ
Now, in the identity sin 𝑋 − sin 𝑌 = 2 cos ( 𝑋+𝑌
) sin ( 𝑋−𝑌
), let 𝑋 = 𝑥 + ℎ, 𝑌 = 𝑥 then
2
2
(
)
ℎ
2 cos ( 2𝑥+ℎ
)
sin
(
)
𝑑𝑦
2
2
= lim
𝑑𝑥 ℎ→0
ℎ
(
)
ℎ sin ( ℎ2 )
= lim cos (𝑥 + ) ℎ
ℎ→0
2
2
)
(
(
)
ℎ
)
sin
(
ℎ
2
= lim cos (𝑥 + ) lim
ℎ
ℎ→0
ℎ→0
2
2
and from theorem 1.4,
sin ( ℎ2 )
ℎ
2
→ 1 as ℎ → 0, which leaves us with
(
)
ℎ
𝑑𝑦
= lim cos (𝑥 + ) = cos 𝑥
𝑑𝑥 ℎ→0
2
Similarly, we may differentiate 𝑥−𝑛 for positive integral 𝑛, and cos 𝑥 (see example sheets)
and also the functions 𝑒𝑥 and ln 𝑥. We summarise a few important derivatives which should be
committed to memory.
1.3.1
Some Important Derivatives
Function
Derivative
𝑥𝑛
𝑛𝑥𝑛−1
sin 𝑥
cos 𝑥
cos 𝑥
−sin 𝑥
𝑥
𝑒
𝑒𝑥
ln 𝑥
1
𝑥
Multiplication by a Scalar and a Sum of two functions
Let 𝑢(𝑥) and 𝑣(𝑥) be functions of 𝑥 whose derivatives are known and let 𝑘 be a constant number.
Theorem 1.8
1.
Then the derivative of the function 𝑘 𝑢(𝑥), is given by
𝑑
𝑑𝑢
(𝑘 𝑢) = 𝑘
𝑑𝑥
𝑑𝑥
2.
Then the derivative of their sum 𝑢(𝑥) + 𝑣(𝑥), is given by
𝑑
𝑑𝑢 𝑑𝑣
(𝑢 + 𝑣) =
+
𝑑𝑥
𝑑𝑥 𝑑𝑥
5
Proof: Using the definition of derivative, let 𝑓 (𝑥) = 𝑢(𝑥) + 𝑣(𝑥)
(
)
(
)
𝑑𝑓
𝑓 (𝑥 + ℎ) − 𝑓 (𝑥)
𝑢(𝑥 + ℎ) + 𝑣(𝑥 + ℎ) − 𝑢(𝑥) − 𝑣(𝑥)
= lim
= lim
ℎ→0
𝑑𝑥 ℎ→0
ℎ
ℎ
(
)
𝑢(𝑥 + ℎ) − 𝑢(𝑥) 𝑣(𝑥 + ℎ) − 𝑣(𝑥)
= lim
+
ℎ→0
ℎ
ℎ
)
(
)
(
𝑣(𝑥 + ℎ) − 𝑣(𝑥)
𝑑𝑢 𝑑𝑣
𝑢(𝑥 + ℎ) − 𝑢(𝑥)
+ lim
=
+
= lim
ℎ→0
ℎ→0
ℎ
ℎ
𝑑𝑥 𝑑𝑥
A function 𝑓 (𝑥) may be the product of two simpler functions, e.g. 𝑓 (𝑥) = (𝑥2 + 1)𝑒𝑥 is a
product of 𝑥2 + 1 and 𝑒𝑥 . Then 𝑓 (𝑥) may be differentiated using the product rule.
Theorem 1.9
Product Rule
Let 𝑓 (𝑥) = 𝑢(𝑥)𝑣(𝑥), where 𝑢(𝑥) and 𝑣(𝑥) are functions of 𝑥. Then
𝑑
𝑑𝑢
𝑑𝑣
(𝑢(𝑥)𝑣(𝑥)) = 𝑣
+𝑢
𝑑𝑥
𝑑𝑥
𝑑𝑥
Proof: First Method. Let 𝑦(𝑥) = 𝑢(𝑥)𝑣(𝑥). As 𝑥 is incremented to 𝑥 + 𝛿𝑥, let 𝑢, 𝑣, 𝑦 change
to 𝑢 + 𝛿𝑢, 𝑣 + 𝛿𝑣, 𝑦 + 𝛿𝑦 then
𝑦(𝑥 + 𝛿𝑥) = 𝑦 + 𝛿𝑦 = (𝑢 + 𝛿𝑢)(𝑣 + 𝛿𝑣)
and so
𝛿𝑦 = (𝑢 + 𝛿𝑢)(𝑣 + 𝛿𝑣) − 𝑢𝑣
i.e.
𝛿𝑦 = 𝑢𝛿𝑣 + 𝑣𝛿𝑢 + 𝛿𝑢𝛿𝑣
Using the definition of derivative,
( )
𝑑𝑦
𝛿𝑦
= lim
𝑑𝑥 𝛿𝑥→0 𝛿𝑥
(
)
𝑢𝛿𝑣 + 𝑣𝛿𝑢 + 𝛿𝑢𝛿𝑣
= lim
𝛿𝑥→0
𝛿𝑥
i.e.
(
)
𝛿𝑣
𝛿𝑢
𝛿𝑣
= lim 𝑢 + 𝑣 + 𝛿𝑢
𝛿𝑥→0
𝛿𝑥
𝛿𝑥
𝛿𝑥
now using the theorems about limits,
(
)
(
)
(
)
𝛿𝑣
𝛿𝑢
𝛿𝑣
= lim 𝑢
+ lim 𝑣
+ lim 𝛿𝑢
𝛿𝑥→0
𝛿𝑥→0
𝛿𝑥→0
𝛿𝑥
𝛿𝑥
𝛿𝑥
(
)
𝛿𝑣
𝛿𝑢
𝛿𝑣
= 𝑢 lim
+ 𝑣 lim
+ lim 𝛿𝑢 lim
𝛿𝑥→0 𝛿𝑥
𝛿𝑥→0 𝛿𝑥
𝛿𝑥→0
𝛿𝑥→0 𝛿𝑥
𝑑𝑣
𝑑𝑢
=𝑢 +𝑣
𝑑𝑥
𝑑𝑥
6
which is the required formula, and completes the proof.
Second method. Again, using the definition of derivative,
(
)
𝑑𝑓
𝑓 (𝑥 + ℎ) − 𝑓 (𝑥)
= lim
𝑑𝑥 ℎ→0
ℎ
)
(
𝑢(𝑥 + ℎ)𝑣(𝑥 + ℎ) − 𝑢(𝑥)𝑣(𝑥)
= lim
ℎ→0
ℎ
)
(
𝑢(𝑥 + ℎ)𝑣(𝑥 + ℎ) − 𝑢(𝑥 + ℎ)𝑣(𝑥) + 𝑢(𝑥 + ℎ)𝑣(𝑥) − 𝑢(𝑥)𝑣(𝑥)
= lim
ℎ→0
ℎ
)
(
))
(
(
𝑢(𝑥 + ℎ) − 𝑢(𝑥)
𝑣(𝑥 + ℎ) − 𝑣(𝑥)
+ 𝑣(𝑥)
= lim 𝑢(𝑥 + ℎ)
ℎ→0
ℎ
ℎ
but the limit of a sum is the sum of the limits
(
(
))
(
(
))
𝑣(𝑥 + ℎ) − 𝑣(𝑥)
𝑢(𝑥 + ℎ) − 𝑢(𝑥)
= lim 𝑢(𝑥 + ℎ)
+ lim 𝑣(𝑥)
ℎ→0
ℎ→0
ℎ
ℎ
and letting ℎ → 0, 𝑢(𝑥 + ℎ) → 𝑢(𝑥), giving
(
)
(
)
𝑣(𝑥 + ℎ) − 𝑣(𝑥)
𝑢(𝑥 + ℎ) − 𝑢(𝑥)
= 𝑢(𝑥) lim
+ 𝑣(𝑥) lim
ℎ→0
ℎ→0
ℎ
ℎ
=𝑢
𝑑𝑢
𝑑𝑣
+𝑣
𝑑𝑥
𝑑𝑥
which is the required formula.
Example 1.10
Let 𝑦 = 𝑥2 sin 𝑥. Let 𝑢 = 𝑥2 and 𝑣 = sin 𝑥. Then
𝑑𝑦
𝑑
𝑑𝑢
𝑑𝑣
=
(𝑢𝑣) = 𝑣
+𝑢
= (sin 𝑥)2𝑥 + 𝑥2 (cos 𝑥)
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
= 2𝑥 sin 𝑥 + 𝑥2 cos 𝑥
Example 1.11
Let 𝑦 = 𝑥ln 𝑥. We can let 𝑢 = 𝑥 and 𝑣 = ln 𝑥. Then
( )
𝑑𝑦
𝑑
𝑑𝑢
𝑑𝑣
1
=
(𝑢𝑣) = 𝑣
+𝑢
= (ln 𝑥)1 + 𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑥
= 1 + ln 𝑥
Suppose we wish to differentiate the function 𝑦 = sin (𝑥2 ). This is a function of a function in
the sense that the output from the function 𝑥2 is the input for the 𝑠𝑖𝑛𝑒 function. Let 𝑢 = 𝑥2 ,
then 𝑦 = sin𝑢, and we now have two simpler functions, each of which we can differentiate
differentiate. The following theorem enables us to differentiate the composite function 𝑦 =
sin (𝑥2 ). A function like this is called a function of a function. In the example, 𝑦 is the function
sin of the function 𝑥2 .
7
Theorem 1.12
Chain Rule or Function of a Function Rule Let 𝑦 = 𝑓 (𝑢) and 𝑢 = 𝑔(𝑥) then the
derivative of the function 𝑦 = 𝑓 (𝑔(𝑥)) is given by
𝑑𝑦
𝑑𝑦 𝑑𝑢
=
𝑑𝑥
𝑑𝑢 𝑑𝑥
Proof: From the definition of derivative
𝑑𝑦
𝛿𝑦
= lim
𝑑𝑥 𝛿𝑥→0 𝛿𝑥
where 𝛿𝑥 represents an increment in 𝑥 and 𝛿𝑦 representing the corresponding increment in 𝑦.
Let the change in 𝑢 corresponding to an increment 𝛿𝑥 in 𝑥 be 𝛿𝑢 then
𝛿𝑦
𝛿𝑦 𝛿𝑢
=
𝛿𝑥
𝛿𝑢 𝛿𝑥
and so
𝑑𝑦
𝛿𝑦 𝛿𝑢
= lim
.
𝑑𝑥 𝛿𝑥→0 𝛿𝑢 𝛿𝑥
Using the theorems on limits,
𝑑𝑦
=
𝑑𝑥
(
𝛿𝑦
lim
𝛿𝑥→0 𝛿𝑢
) (
𝛿𝑢
lim
𝛿𝑥→0 𝛿𝑥
)
.
𝑑𝑦 𝑑𝑢
𝑑𝑢 𝑑𝑥
which is the derivative of 𝑦 with respect to 𝑢 multiplied by the derivative of 𝑢 with respect to
𝑥.
=
Example 1.13
Differentiate 𝑦 = sin (𝑥2 ). Let 𝑢 = 𝑥2 then 𝑦 = sin𝑢 and
𝑑𝑦
= cos 𝑢,
𝑑𝑢
𝑑𝑢
= 2𝑥.
𝑑𝑥
Hence, using the chain rule
𝑑𝑦
𝑑𝑦 𝑑𝑢
=
= (cos 𝑢)2𝑥 = 2𝑥 cos 𝑥2
𝑑𝑥
𝑑𝑢 𝑑𝑥
With practice it is possible to use the chain rule without explicitly making the substitution.
Example 1.14
Let 𝑦 = 𝑒sin 𝑥 . Let 𝑢 = sin 𝑥, then 𝑦 = 𝑒𝑢 . Now, applying the chain rule:
𝑑𝑦
𝑑𝑦 𝑑𝑢
=
= 𝑒𝑢 cos 𝑥 = cos 𝑥 𝑒sin 𝑥
𝑑𝑥
𝑑𝑢 𝑑𝑥
Example 1.15
Let 𝑦 = ln (𝑥 + 3). Let 𝑢 = 𝑥 + 3, then 𝑦 = ln 𝑢. Now, applying the chain rule:
𝑑𝑦
𝑑𝑦 𝑑𝑢
1
1
=
= 1=
𝑑𝑥
𝑑𝑢 𝑑𝑥
𝑢
𝑥+3
8
Example 1.16
Let 𝑦 =
1
𝑤
= 𝑤−1 for any function 𝑤 of 𝑥. Now, applying the chain rule:
𝑑𝑦
𝑑𝑦 𝑑𝑤
𝑑𝑤
1 𝑑𝑤
=
= −𝑤−2
=− 2
𝑑𝑥
𝑑𝑤 𝑑𝑥
𝑑𝑥
𝑤 𝑑𝑥
Theorem 1.17
Quotient Rule. Let
𝑢
= 𝑢𝑣 −1
𝑣
for any functions 𝑢 and 𝑣 of 𝑥. Let 𝑤 = 𝑣 −1 and apply the product rule:
(
)
𝑑𝑦
𝑑𝑢
𝑑𝑤
1 𝑑𝑣
−1 𝑑𝑢
=𝑤
+𝑢
=𝑣
+𝑢 − 2
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑣 𝑑𝑥
𝑦=
=
𝑑𝑣
𝑣 𝑑𝑢
− 𝑢 𝑑𝑥
𝑑𝑥
𝑣2
This is known as the quotient rule.
Example 1.18
Let 𝑦 =
𝑒𝑥
. So, using the quotient rule:
𝑥
𝑣 𝑑𝑢 − 𝑢 𝑑𝑣
𝑥𝑒𝑥 − 𝑒𝑥
𝑑𝑦
= 𝑑𝑥 2 𝑑𝑥 =
=
𝑑𝑥
𝑣
𝑥2
Example 1.19
Let 𝑦 = tan 𝑥 =
(
𝑥−1
𝑥2
)
𝑒𝑥
sin 𝑥
. So, we using the quotient rule:
cos 𝑥
𝑣 𝑑𝑢 − 𝑢 𝑑𝑣
cos 𝑥 cos 𝑥 − sin 𝑥 (−cos 𝑥)
𝑑𝑦
= 𝑑𝑥 2 𝑑𝑥 =
𝑑𝑥
𝑣
cos2 𝑥
1
=
= sec2 𝑥
2
cos 𝑥
Using the basic functions, and the rules for differentiating products, quotients and function–
of–a–function, it is possible to differentiate a wide range of functions. Two last examples should
suffice:
2
𝑒𝑥
2
Example 1.20 Let 𝑦 =
. This is a quotient, so we use that rule, but along the way 𝑒𝑥
sin 𝑥
needs to be differentiated using the chain rule.
2
2
𝑣 𝑑𝑢 − 𝑢 𝑑𝑣
𝑑𝑦
(sin 𝑥)(2𝑥𝑒𝑥 ) − (𝑒𝑥 )(cos 𝑥)
= 𝑑𝑥 2 𝑑𝑥 =
𝑑𝑥
𝑣
sin2 𝑥
2
𝑒𝑥
= (2𝑥sin 𝑥 − cos 𝑥) 2
sin 𝑥
Example 1.21 Let 𝑦 = (1 + cos 𝑥2 )6 . This is a function of a function of a function: let 𝑢 =
1 + cos 𝑥2 then 𝑦 = 𝑢6 may be differentiated with respect to 𝑢. To differentiate 𝑢 = 1 + cos 𝑥2
we may use the chain rule again to differentiate 𝑢 w.r.t 𝑥. So let 𝑣 = 𝑥2 then 𝑢 = 1 + cos 𝑣.
The formula for the derivative becomes
𝑑𝑦
𝑑𝑦 𝑑𝑢 𝑑𝑣
=
𝑑𝑥
𝑑𝑢 𝑑𝑣 𝑑𝑥
9
Note: this is the product of all the functions, differentiated with respect to their argument and
multiplied together (justifying the term chain rule). Hence,
𝑑𝑦
= 6𝑢5 (−sin 𝑣) 2𝑥
𝑑𝑥
= 6(1 + cos 𝑥2 )5 (−2𝑥 sin 𝑥2 ) = −12𝑥 sin 𝑥2 (1 + cos 𝑥2 )5
1.4
Implicit Differentiation
Let 𝑦 = 𝑓 (𝑥) then 𝑦 is called an explicit function of 𝑥, e.g. 𝑦 = 𝑥2 . Substituting a value of 𝑥
gives a value of 𝑦. However, consider the example
𝑥2 + 𝑦 2 = 1.
which represents the equation of a circle. In this case substituting a value of 𝑥 such that
−1 ≤ 𝑥 ≤ +1 merely gives a value for 𝑦 2 . To find 𝑦 further operations are required, including
taking the√square root. Here, 𝑦 is not a function of 𝑥, since rearranging the expression to
get 𝑦 = ± 1 − 𝑥2 , we may have two values for 𝑦, or 1 value or even none ( a function must
give exactly one value). The general case is often written 𝑓 (𝑥, 𝑦) = 0 and is called an implicit
relation between 𝑥 and 𝑦. In order to differentiate an implicit expression, we proceed as follows.
Example 1.22
In chapter 2 we obtained the trigonometric identity sin 2𝑥 = 2sin 𝑥 cos 𝑥.
Suppose we differentiate both sides (the left is a function of a function, while the right is a
product):
2 cos 2𝑥 = 2(sin 𝑥)(−sin 𝑥) + (cos 𝑥)(cos 𝑥) = 2(cos2 𝑥 − sin2 𝑥)
and cos2 𝑥−sin2 𝑥 is the formula for cos2𝑥, so, differentiating an identity also gives an identity.
We can also differentiate:
)
𝑑 ( 2
𝑑
(1)
𝑥 + 𝑦2 =
𝑑𝑥
𝑑𝑥
𝑑 ( 2)
𝑑 ( 2)
𝑥 +
𝑦 =0
𝑑𝑥
𝑑𝑥
𝑑
To obtain 𝑑𝑥
(𝑦 2 ) we use the chain rule: let 𝑢 = 𝑦 2 , then
𝑑𝑢
𝑑𝑢 𝑑𝑦
𝑑𝑦
=
= 2𝑦 .
𝑑𝑥
𝑑𝑦 𝑑𝑥
𝑑𝑥
Hence
2𝑥 + 2𝑦
𝑑𝑦
=0
𝑑𝑥
and rearranging this gives:
𝑑𝑦
𝑥
=−
𝑑𝑥
𝑦
Note that the right hand side contains 𝑦, and this may be obtained by solving the implicit
relation for 𝑦, if required. Implicit differentiation is very useful in finding the derivatives of
inverse functions.
10
1.5
Example
Let 𝑦 = tan−1 𝑥, the inverse tan function (and not
1
).
tan 𝑥
We can rearrange the formula so as
to eliminate the inverse function in favour of 𝑡𝑎𝑛 by taking the tangent of both sides, to obtain
𝑥 = tan 𝑦 and then differentiate both sides:
𝑑
𝑑
(𝑥)=
(tan 𝑦)
𝑑𝑥
𝑑𝑥
𝑑𝑦
𝑑𝑥
Now, from the identity sec2 𝑦 = 1 + tan2 𝑦 = 1 + 𝑥2 , we get
1 = sec2 𝑦
𝑑𝑦
1
=
𝑑𝑥
1 + 𝑥2
which is the formula for the derivative of tan−1 𝑥.
We end this section on differentiation by a summary of results. Some of these have been
proved, or solved in examples. Others are on exercise sheets.
Function
Derivative
𝑥𝑛
𝑛𝑥𝑛−1
sin 𝑥
cos 𝑥
cos 𝑥
−sin 𝑥
tan 𝑥
sec2 𝑥
cot 𝑥
−cosec2 𝑥
sec 𝑥
sec 𝑥 tan 𝑥
cosec 𝑥
−cosec 𝑥 cot 𝑥
𝑥
𝑒
𝑒𝑥
ln 𝑥
1
𝑥
sin−1 𝑥
√
cos−1 𝑥
tan−1 𝑥
2
1
1 − 𝑥2
−1
√
1 − 𝑥2
1
1 + 𝑥2
Applications of the Derivative
𝑑𝑦
The derivative has been defined as the gradient of a curve. In other words, 𝑑𝑥
represents the
rate of change of 𝑦 with respect to 𝑥. As an example, if 𝑦 represents distance, and 𝑥 time,
this becomes the rate of change of distance with respect to time, i.e. speed. Note also, if we
differentiate again, we measure the rate of change of speed, i.e. acceleration.
𝑑𝑓
Suppose we have a function 𝑓 (𝑥). We may interpret 𝑑𝑥
= 0 as follows.
11
𝑓 (𝑥)
𝑑𝑓
= 0, the gradient of the graph is zero and the tangent to the curve is horizontal.
When 𝑑𝑥
A point at which this happens is called a stationary point or critical point. There are three
possibilities. The diagram shows a local maximum and the gradient is changing from positive
to negative as we move from left to right through the stationary point. Similarly, we may draw
a local minimum and in this case the gradient changes from negative to positive as we move
from left to right through the stationary point. Such points are called turning points. The third
possibility is that the sign of the gradient remains the same (either positive or negative) either
side of the stationary point and this is called a point of inflection. An example is provided by
the curve of 𝑦 = 𝑥3 at the point 𝑥 = 0, the gradient is zero there but the gradient is positive
everywhere else.
2.1
Example
Find the maximum value of the function 𝑦 = 2 + 3𝑥 − 𝑥2
Solution: Differentiating,
𝑑𝑦
= 3 − 2𝑥 = 0
𝑑𝑥
which gives 𝑥 = 1.5. Substituting this into the equation, we get 𝑦 = 4.25. To see whether it is
a maximum or a minimum, we may calculate the gradient either side of 𝑥 = 1.5,
𝑓 ′ (1.4) = 0.2 > 0
𝑓 ′ (1.6) = −0.2 < 0
so we deduce that 4.25 is the maximum.
𝑑𝑦
A function may have several values of 𝑥 where 𝑑𝑥
= 0, for example, sin 𝑥 has many (in fact
infinitely many) maxima and minima.
There is an alternative way of classifying stationary points, which involves finding the second
derivative of the function and evaluating it at the stationary point.
Theorem 2.1
Second derivative test. Let 𝑦 = 𝑓 (𝑥) and let
if
𝑑2 𝑦
>0
𝑑𝑥2
12
𝑑𝑦
𝑑𝑥
= 0 at the point 𝑥 = 𝑥0 . Then 1.
at 𝑥 = 𝑥0 then the stationary point is a minimum.
2. if
𝑑2 𝑦
<0
𝑑𝑥2
at 𝑥 = 𝑥0 then the stationary point is a maximum.
3. if
𝑑2 𝑦
=0
𝑑𝑥2
at 𝑥 = 𝑥0 then the test fails and further investigation is needed.
Example 2.2
Find and classify the turning points of the function 𝑦 = 𝑥3 − 3𝑥 + 1.
Solution: Differentiating,
𝑑𝑦
= 3𝑥2 − 3 = 0
𝑑𝑥
so there are two turning points at 𝑥 = 1 and 𝑥 = −1. Using the second derivative test:
𝑑2 𝑦
= 6𝑥
𝑑𝑥2
2
𝑑 𝑦
If 𝑥 = +1 then 𝑑𝑥
2 = +6 which is positive, so we have a minimum.
𝑑2 𝑦
If 𝑥 = −1 then 𝑑𝑥2 = −6 which is negative, so we have a maximum.
Let 𝑥 = 0 then 𝑦 = 1. This gives sufficient information to sketch the curve.
13
3
1
1
−1
−1
More examples of maxima and minima can be found on the problem sheets.
14