Introduction
Reverse chain rule
Usage
Test
Examples
f ′ (x)/f (x)
Test
REVERSE CHAIN RULE
CALCULUS 7
INU0115/515 (M ATHS 2)
Dr Adrian Jannetta MIMA CMath FRAS
Reverse Chain Rule
1 / 12
Adrian Jannetta
Introduction
Reverse chain rule
Usage
Test
Examples
f ′ (x)/f (x)
Test
Reversing the chain rule
In differentiation the chain rule is used to get the derivative of a composite
function. For example
d 2x
(e ) = 2e2x
dx
Integration is the reverse of differentiation. Therefore, we could reverse this
statement and write:
Z
2e2x dx = e2x
(Ignore the constant for a moment). Dividing both sides by 2 we find
Z
e2x
e2x dx =
2
Similarly, consider
d
dx
(1)
[sin(3x + 2)] = 3 cos(3x + 2). Reversing this process:
Z
3 cos(3x + 2) dx = sin(3x + 2)
Therefore:
Z
sin(3x + 2)
(2)
3
Can you a pattern for directly integrating the expressions given in (1) and (2)?
Reverse Chain Rule
cos(3x + 2) dx =
2 / 12
Adrian Jannetta
Introduction
Reverse chain rule
Usage
Test
Examples
f ′ (x)/f (x)
Test
Reversing the chain rule
Here’s another example of reversing the chain rule.
d
(2x − 5)3 = (3)(2)(2x − 5)2
dx
In words this process is ‘Subtract 1 from the power. Multiply by the old power and multiply
by the derivative of the inner function’.
Since integration is the reverse of differentiation:
Z
(3)(2)(2x − 5)2 dx = (2x − 5)3
(Again, let’s ignore the constant C for a moment). Dividing both sides by the factor (3)(2) we
get
Z
(2x − 5)3
(2x − 5)2 dx =
(3)(2)
Notice that the (3) is the power and the (2) is the derivative of the inner function (2x − 5)
Z
(2x − 5)3
(2x − 5)2 dx =
6
The first term on the RHS is equivalent in words to ‘Add 1 to the power. Divide by the new
power and divide by the derivative of the inner function’. It is the reverse of applying the
chain rule.
Reverse Chain Rule
3 / 12
Adrian Jannetta
Introduction
Reverse chain rule
Usage
Test
Examples
f ′ (x)/f (x)
Test
Reverse Chain Rule
The examples just seen illustrate a simple procedure for integrating
composite functions. It is the reverse of how we’d carry out the chain
rule in differentiation.
In words, the general procedure doing reverse chain rule is
✗
✔
✖
✕
‘Integrate the outer function. Divide by
the derivative of the inner function’
Here some examples. We’ll include the constant from now on!
Z
− cos 10x
+C
sin 10x dx =
10
Z
Reverse Chain Rule
e−4x dx =
e−4x
+C
−4
4 / 12
Adrian Jannetta
Introduction
Reverse chain rule
Usage
Test
Examples
f ′ (x)/f (x)
Test
Reverse chain rule
Use the reverse chain rule to integrate
Z
1
dx
(3 − 5x)4
First, write the integral in a more suitable form
Z
(3 − 5x)−4 dx
Now integrate. Use the rule for integrating powers (in this case, like x−4 );
‘Add 1 to the power. Divide by the new power and divide by the
derivative of the inner function’.
Z
(3 − 5x)−3
(3 − 5x)−3
(3 − 5x)−4 dx =
+C =
+C
(−3)(−5)
15
It can be written without the negative power like this:
Z
1
1
dx =
+C
∴
4
(3 − 5x)
15(3 − 5x)3
Reverse Chain Rule
5 / 12
Adrian Jannetta
Introduction
Reverse chain rule
Usage
Test
Examples
f ′ (x)/f (x)
Test
When can reverse chain rule be used?
The reverse chain rule can only be used when the inner function is
linear, which means it has the form mx + c (where m and c are constants).
✔
✗
We cannot use the reverse chain rule
when the inner function is NOT LINEAR!
✖
For example, we could not use it to find
Z
✕
(1 + x2 )5 dx because the inner
function 1 + x2 is not linear.
Therefore
Z
(1 + x2 )5 dx 6=
(1 + x2 )6
+C
6(2x)
(You can check this by finding the derivative of the RHS - it will not be
equal to the LHS!)
Reverse Chain Rule
6 / 12
Adrian Jannetta
Introduction
Reverse chain rule
Usage
Test
f ′ (x)/f (x)
Examples
Test
Test yourself
Which of the following can be integrated with the reverse chain rule?
1
Z
(1 − 3x) dx
5
Z
2
Z
2e−x dx
6
Z
3
Z
7
Z
p
1 + 2x dx
4
Z
1
dx
1 − x2
1
dx
1 − 3x
8
Z
2x sin x dx
4
p
sin x dx
cos(3x − π) dx
Answers...
1
Yes
3
No
5
No
7
Yes
2
Yes
4
Yes
6
Yes
8
No
Reverse Chain Rule
7 / 12
Adrian Jannetta
Introduction
Reverse chain rule
Usage
Test
Examples
f ′ (x)/f (x)
Test
Further examples
Rational power
Z
p
Find
2x + 5 dx with the reverse chain rule.
First rewrite without the square-root:
Z
3
1
2
(2x + 5) dx
=
(2x + 5) 2
( 23 )(2)
+C
3
=
=
(2x + 5) 2
+C
3
p
(2x + 5)3
+C
3
You can always check the answer by differentiating it! You should end up
with the original integrand.
Reverse Chain Rule
8 / 12
Adrian Jannetta
Introduction
Reverse chain rule
Usage
Test
Examples
f ′ (x)/f (x)
Test
A definite integral
Z1
(2x + 1)3 dx
Evaluate
0
Using the reverse chain rule we see that
Z1
(2x + 1)3 dx
=
1
(2x + 1)4
(4)(2) 0
1
4 1
8 (2x + 1) 0
1
4
4
8 3 −1
=
10
=
0
=
=
1
8
× (81 − 1)
Looking forward...
The reverse chain rule is a useful shortcut for simple situations. A related
technique, called integration by substitution, will be demonstrated in the
next lecture.
Reverse Chain Rule
9 / 12
Adrian Jannetta
Introduction
Reverse chain rule
Usage
Test
Examples
f ′ (x)/f (x)
Test
Integration of f ′ (x)/f (x)
Here is a result that we saw during our study of differentiation:
f ′ (x)
d ln f (x) =
dx
f (x)
It’s a consequence of differentiating logarithm functions using the chain
rule.
Naturally, we can reverse this to produce an integration result:
Z
f ′ (x)
dx = ln f (x) + C
f (x)
(3)
This is a result we should learn to recognise when it appears!
Reverse Chain Rule
10 / 12
Adrian Jannetta
Introduction
Reverse chain rule
Usage
Test
Examples
f ′ (x)/f (x)
Test
Integration of f ′ (x)/f (x)
Find
Z
2x
dx.
x2 − 9
This integral matches the form of equation (3) exactly.
Therefore:
Z
2x
dx = ln x2 − 9 + C
x2 − 9
Integration of f ′ (x)/f (x)
Find
Z
4 sin x
dx.
1 + cos x
This integral almost fits the form of equation (3). We can rewrite it to make it the same:
Z
Z
4 sin x
− sin x
dx = −4
dx
1 + cos x
1 + cos x
We took out a factor of −4 so that the numerator is now the derivative of the denominator.
Therefore:
Reverse Chain Rule
Z
4 sin x
dx = − 4 ln |1 + cos x| + C
1 + cos x
11 / 12
Adrian Jannetta
Introduction
Reverse chain rule
Usage
Test
Examples
f ′ (x)/f (x)
Test
Test yourself
Use of reverse chain rule
Use the reverse chain rule to find these integrals:
1
Z
(1 − 3x)4 dx
2
Z
2e−x dx
3
Z
4
Z
1
dx
5 + 2x
2 cos( 31 x − π) dx
5
Z
6
Z
7
Z
5
1
6
π
9
sin 3x dx
0
4
0
1
dx
p
1 + 2x
cos x
dx
sin x
Answers...
1
5
1 − 15 (1 − 3x) + C
−x
2 −2e + C
3
1
2
4
6 sin( 13 x − π) + C
ln |5 + 2x| + C
Reverse Chain Rule
6 2
7 ln |sin x| + C
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Adrian Jannetta