Extending the table

of derivatives
In this unit we continue to build up The Table of Derivatives using rules described in other
units.

In order to master the techniques explained here it is vital that you undertake plenty of practice
exercises so that they become second nature.

After reading this text, and/or viewing the video tutorial on this topic, you should be able to:

• differentiate trigonometric functions

• differentiate inverse trigonometric functions

• differentiate ax .

Contents
1. Introduction 2
2. Examples 2

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All references to © mathtutor on subsequent pages should be read as © skillbank solutions limited 2010
1. Introduction
In this unit we construct several new entries for the Table of Derivatives using standard rules
and results obtained previously. The table we will construct is shown here:

df
function f (x) derivative or f ′ (x)
dx
sin x
tan x = cos x
sec2 x
sec x = cos1 x sec x tan x
x
cot x = cos
sin x
−cosec 2 x
cosec x = sin1 x −cosec x cot x
tan mx m sec2 mx
sec mx m sec mx tan mx
cot mx −m cosec2 mx
cosec mx −m cosec mx cot mx
sin−1 x √ 1
1−x2
1
cos−1 x − √1−x 2

1
tan−1 x 1+x2

ax x
a ln a

Most of these new entries are derived in the following examples.

2. Examples
Example 1
sin x
Suppose we wish to differentiate y = tan x. Note that we can write this as y = tan x = cos x
.
Because this is a quotient we can use the quotient rule to perform the differentiation.
The quotient rule states:
u dy v du − u dv
if y= then = dx 2 dx
v dx v
sin x
So to differentiate y = cos x
we take u = sin x and v = cos x. Then
du dv
= cos x and = − sin x
dx dx
Then, applying the quotient rule:
du dv
dy v dx − u dx
=
dx v2
cos x(cos x) − sin x(− sin x)
=
(cos x)2
cos2 x + sin2 x
=
cos2 x

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A key trigonometric identity states that cos2 x + sin2 x = 1 and so this simplifies to
dy 1
=
dx cos2 x
which can also be written as sec2 x. So the derivative of tan x is sec2 x.
Example 2
Suppose we wish to differentiate y = sec x.
Note that we can write this as y = sec x = cos1 x . Because this is a quotient we can again use the
quotient rule to perform the differentiation.
The quotient rule states:

u dy v du
dx
dv
− u dx
if y= then =
v dx v2
1
So to differentiate y = cos x
we take u = 1 and v = cos x. Then

du dv
=0 and = − sin x
dx dx
Then, applying the quotient rule:

dy v du
dx
dv
− u dx
=
dx v2
cos x(0) − 1(− sin x)
=
(cos x)2
sin x
=
cos2 x
this can be written as
dy 1 sin x
= = sec x tan x
dx cos x cos x
So the derivative of sec x is sec x tan x.
Example 3
Suppose we wish to differentiate y = tan mx where m is a constant.
We make a substitution to simplify this function. Suppose we let u = mx so that y = tan u.
dy
We can use the chain rule to find dx :
dy dy du
= ×
dx du dx
In this case, since u = mx then du
dx
= m.
dy
Since y = tan u, we have just seen that du = sec2 u,
So, the chain rule gives
dy
= sec2 u × m
dx
= m sec2 mx

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Example 4
Suppose we wish to differentiate y = cosec mx where m is a constant.
We make a substitution to simplify this function. Suppose we let u = mx so that y = cosec u.
dy
We can again use the chain rule to find dx : In this case, since u = mx then du
dx
= m.
dy
Since y = cosec u, we note from the Table on page 2 that du = −cosec u cot u,
So, the chain rule gives
dy
= −cosec u cot u × m
dx
= −m cosec mx cot mx
Example 5
Suppose we wish to differentiate y = sin−1 x.
We proceed by rewriting this as sin y = x. We then differentiate both sides with respect to x:
d d
(sin y) = (x)
dx dx
The right hand side is straightforward because the derivative of x with respect to x is just 1.
The left hand side needs more care because we need to differentiate a function of y, that is sin y,
with respect to x. We do this implicitly as follows:
d d dy
(sin y) = (sin y) ×
dx dy dx
dy
= cos y
dx
(If necessary you should refer to the unit on implicit differentiation in order to understand this
process.) Putting these results together we find
dy
cos y =1
dx
Therefore
dy 1
=
dx cos y
We now try to write the right hand side in terms of x. We can do this using the identity
cos2 y + sin2 y = 1
so that q
cos y = 1 − sin2 y
We take only the positive square root. This is because studying the graph of y = sin−1 x shows
dy
that its gradient, and hence dx , is positive.
So
dy 1 1
= =p
dx cos y 1 − sin2 y
Finally, recall that sin y = x so that
dy 1
=√
dx 1 − x2
So the derivative of y = sin−1 x is √ 1 .
1−x2

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August 10, 2004
A similar argument can be used to find the derivative of y = cos−1 x and this is left as an
exercise.
Example 6
dy
Suppose y = tan−1 x and we wish to find dx . We proceed by rewriting this as tan y = x. We
then differentiate both sides with respect to x:
d d
(tan y) = (x)
dx dx
The right hand side is straightforward because the derivative of x with respect to x is just 1.
The left hand side needs more care because we need to differentiate a function of y, that is tan y,
with respect to x. We do this implicitly as follows:
d d dy
(tan y) = (tan y) ×
dx dy dx
dy
= sec2 y
dx
Putting these results together we find
dy
sec2 y =1
dx
Therefore
dy 1
=
dx sec2 y
1
=
1 + tan2 y
1
=
1 + x2
Here we have made use of the trigonometric identity 1 + tan2 y = sec2 y. So the derivative of
1
y = tan−1 x is 1+x 2.

Example 7
Suppose we want to differentiate y = ax . We proceed by taking logarithms of both sides:
ln y = ln ax = x ln a
using the laws of logarithms.
Differentiating both sides with respect to x gives
d
ln y = ln a
dx
since ln a is a constant. Now, using the chain rule,
d d dy
ln y = ln y ×
dx dy dx
so
1 dy
= ln a
y dx
Then
dy
= y ln a = ax ln a
dx

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Exercises

1. Use the extended table of derivatives in Section 1 to find the derivative of each of the
following:

a) tan 3x b) cos−1 x c) 5x d) cot 5x

x
e) cot 3x f) 2 g) cosec4x h) tan−1 x
 
1
i) sec 3x j) tan 4x k) 1x l) sec x
2
cos x
2. By writing cot x = and using the quotient rule find the derivative of cot x.
sin x
1
3. By writing cosecx = and using the quotient rule find the derivative of cosecx.
sin x
4. By implicitly differentiating cos y = x determine the derivative of cos−1 x.
−1 x
 
5. Use the chain rule to find the derivative of tan .
a

Answers
1
1. a) 3 sec2 3x b) − √ c) 5x ln 5 d) −5cosec2 5x e) −3cosec2 3x
1 − x2
1
f) 2x ln 2
g) −4cosec4x cot 4x h) i) 3 sec 3x tan 3x j) 4 sec2 4x
    1 + x2
1 1 1
k) 0 l) sec x tan x
2 2 2
2. −cosec2 x

3. −cosecx cot x
1
4. − √
1 − x2
a
5. 2
a + x2

mathtutor project: 6
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