Transcript C3.6 Differentiation

A-Level Maths:
Core 3
for Edexcel
C3.6 Differentiation
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The chain rule
The chain rule
Contents
The relationship between
dy
dx
and
dx
dy
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
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Review of differentiation
So far, we have used differentiation to find the gradients of
functions made up of a sum of multiples of powers of x. We
found that:
If y = x n then
dy
= nx n 1
dx
and when xn is preceded by a constant multiplier k we have:
dy
If y = kx then
= knx n1
dx
n
Also:
dy
If y = f ( x ) ± g( x ) then
= f '( x ) ± g '( x )
dx
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Review of differentiation
We will now look at how to differentiate exponential, logarithmic
and trigonometric functions.
We will also look at techniques that can be used to
differentiate:
Compound functions of the form f(g(x)). For example:
x 1
2
3 x 2
e
sin( x3 )
Products of the form f(x) × g(x), such as:
x ln x
xe2 x
f ( x)
Quotients of the form
, such as:
g( x )
3 x +1
e2 x
x2  1
sin x
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3 x2 cos x
x2
ln x
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The chain rule
The chain rule is used to differentiate composite functions.
For instance, suppose we want to differentiate y = (2x + 1)3 with
respect to x.
One way to do this is to expand (2x + 1)3 and differentiate it
term by term.
Using the binomial theorem:
(2 x +1)3 = 8 x3 + 3(4 x 2 ) + 3(2 x ) +1
= 8 x3 +12x2 + 6 x +1
Differentiating with respect to x:
dy
= 24 x 2 + 24 x + 6
dx
= 6(4 x 2 + 4 x +1)
= 6(2 x +1)2
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The chain rule
Another approach is to use the substitution u = 2x + 1 so that
we can write y = (2x + 1)3 as y = u3.
The chain rule states that:
If y is a function of u and u is a function of x, then
dy dy du
=
×
dx du dx
y = u3
dy
= 3u 2
du
So if
where
u = 2x + 1,
du
=2
dx
dy dy du
Using the chain rule:
=
×
= 3u 2 × 2
dx du dx
= 6u 2
= 6(2 x +1)2
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The chain rule
Use the chain rule to differentiate
y = 3 x2  5 with respect to x.
Let
y =u
1
2
u = 3x2 – 5
where
dy 1  21
= 2u
du
du
= 6x
dx
dy dy du 1  21
Using the chain rule,
=
×
= 2 u × 6x
dx du dx
= 3 xu
 21
 21
= 3 x(3 x  5)
2
=
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3x
3 x2  5
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The chain rule
dy
2
Find
given that y =
.
3 4
dx
(7  x )
2
where u = 7 – x3
Let
y = 4 = 2u 4
u
dy
du
= 8u 5
= 3 x 2
du
dx
dy dy du
Using the chain rule:
=
×
= 8u 5 × 3 x2
dx du dx
= 24 x2u 5
= 24 x 2 (7  x 3 )5
24 x 2
=
(7  x 3 )5
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The chain rule using function notation
With practice some of the steps in the chain rule can be done
mentally.
Suppose we have a composite function
y = g(f(x))
If we let
y = g(u)
where
u = f(x)
dy
du
=
g
'(
u
)
= f '( x )
then
and
du
dx
dy dy du
Using the chain rule:
=
×
= g '(u )× f '( x)
dx du dx
But u = f(x) so
dy
= g '(f ( x ))× f '( x )
If y = g(f(x)) then
dx
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The chain rule
All of the composite functions we have looked at so far have
been of the form y = (f(x))n.
In general, using the chain rule,
If y = (f
(x))n
dy
= n(f ( x ))n 1 × f '( x )
then
dx
If we use
to represent f (x) and
write this rule more visually as:
y=
n
dy

=n
dy
to represent f ’(x) we can
n 1
Find the equation of the tangent to the
curve y = (x4 – 3)3 at the point (1, –8).
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The chain rule
y=
(x4
–
3)3

dy
d 4
4
2
= 3( x  3) × ( x  3)
dx
dx
= 3( x 4  3)2 × 4 x3
= 12 x 3 ( x 4  3)2
When x = 1,
dy
= 12(1  3)2 = 48
dx
Using y – y1 = m(x – x1) the equation of the tangent at the point
(1, –8) is:
y + 8 = 48(x – 1)
y = 48x – 48 – 8
y = 48x – 56
y = 8(6x – 7)
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dy
and
The relationship between dx
dx
dy
The chain rule
Contents
The relationship between
dy
dx
and
dx
dy
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
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dy
and
The relationship between dx
dx
dy
Suppose we are given x as a function of y instead of y as a
function of x. For instance,
x = 4y2
dx
We can find dy by differentiating with respect to y:
dx
= 8y
dy
Using the chain rule we can write
dy 1
dy dx
= dx
× = 1, from which we get:
dx dy
dx dy
dx
So by the above result, if
= 8y then
dy
dy
1
=
dx 8 y
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dy
and
The relationship between dx
dx
dy
Find the gradient of the curve with equation
x = 2y3 – 3y – 7 at the point (3, 2).
x = 2y3 – 3y – 7
dx
= 6 y2  3
dy
At the point (3, 2), y = 2:
dx
= 6(2)2  3 = 21
dy
dy 1
= dx
We can now find the gradient using the fact that
dx dy
dy 1
=
dx 21
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Differentiating inverse functions
dy 1
The result
= dx is particularly useful for differentiating
dx dy
inverse functions. For example:
d
(sin1 x ), writing your answer in terms of x.
Find
dy
Let y = sin–1 x so
x = sin y
dx
= cos y
dy
Using the identity
cos2y = 1 – sin2y
1
dy
1
=
=
dx cos y
1  sin2 y
But sin y = x so
d
1
1
(sin x ) =
dy
1  x2
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Differentiating ex and related functions
The chain rule
Contents
The relationship between
dy
dx
and
dx
dy
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
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The derivative of ex
A special property of the exponential function ex is that
dy
If y = e then
= ex
dx
x
From this, it follows that
If y = ke x then
dy
= ke x
dx
where k is a constant.
For example,
if y = 4ex – x3
dy
= 4e x  3 x 2
dx
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Functions of the form ekx
Suppose we are asked to differentiate a function of the form ekx,
where k is a constant. For example,
Differentiate y = e5x with respect to x.
y = eu
Let
dy
= eu
du
where
u = 5x
du
=5
dx
dy dy du
Using the chain rule:
=
×
= eu ×5
dx du dx
= 5eu
= 5e5 x
In practice, we wouldn’t need to include this much working.
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Functions of the form ekx
We would just remember that in general,
If y = e kx then
dy
= ke kx
dx
For example,
d 7x
( e ) = 7e 7 x
dx
d 2 x
(e ) = 2e 2 x
dx
x
3
d
e
(e ) =
dx
3
We can use the chain rule to extend this to any function of the
form ef(x).
x
3
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Functions of the form ef(x)
If y = ef(x) then we can let
y = eu
Let
where
dy
= eu
du
then
Using the chain rule:
So in general,
u = f(x)
du
= f '( x )
dx
dy dy du
=
×
= eu × f '( x )
dx du dx
= f '( x )ef ( x )
If y = e
f ( x)
dy
then
= f '( x )ef ( x )
dx
In words, to differentiate an expression of the form y = ef(x) we
multiply it by the derivative of the exponent.
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Functions of the form ef(x)
Using
to represent f(x) and
y=e
to represent f ’(x):
dy

=
dx
e
For example,
dy 5 x  4
(e
) = 5e5 x  4
dx
dy 3  x
3 x
( e ) = e
dx
dy
x2  9
x2  9
x2  9
= 10 xe
(5e
) = 5 × 2 xe
dx
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Differentiating ln x and related functions
The chain rule
Contents
The relationship between
dy
dx
and
dx
dy
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
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The derivative of ln x
Remember, ln x is the inverse of ex.
So, if
y = ln x
then
x = ey
Differentiating with respect to y gives:
dx
= ey
dy
dy 1
1
= dx = y
dx dy e
But
ey
= x so,
dy 1
=
dx x
dy 1
If y = ln x then
=
dx x
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Functions of the form ln kx
Suppose we want to differentiate a function of the form ln kx,
where k is a constant. For example:
Differentiate y = ln 3x with respect to x.
Let
y = ln u
where
u = 3x
dy 1
=
du u
du
=3
dx
dy dy du 3
Using the chain rule:
=
×
=
dx du dx u
3
=
3x
1
=
x
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Functions of the form ln kx
When functions of the form ln kx are differentiated, the k’s will
always cancel out, so in general,
dy 1
If y = ln kx then
=
dx x
We can use the chain rule to extend to functions of the more
general form y = ln f(x).
y = ln u
Let
where
u = f(x)
du
dy 1
=
= f '( x )
then
du u
dx
Using the chain rule:
dy dy du f '( x )
=
×
=
dx du dx
u
f '( x)
=
f ( x)
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Functions of the form ln (f(x))
In general, using the chain rule
dy f '( x)
If y = ln f ( x) then
=
dx f ( x )
Using
to represent f(x) and
y = ln
For example,
to represent f ’(x):
dy

=
dy
d
7
ln(7 x  4) =
7x  4
dx
2
d
9
x
ln(3 x3 + 8) = 3
dx
3x + 8
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Functions of the form ln (f(x))
In some cases we can use the laws of logarithms to simplify a
logarithmic function before differentiating it.
Remember that,
ln (ab) = ln a + ln b
a
ln = ln a  ln b
b
ln an = n ln a
x
Differentiate y = ln
with respect to x.
2
x
= ln x  ln2
y = ln
2
1
2
= ln x  ln2
= 21 ln x  ln2
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Functions of the form ln (f(x))
y = 21 ln x  ln2

1
dy 1 1
= × =
dx 2 x 2x
ln 2 is a constant and so it disappears when we differentiate.
x
If we had tried to differentiate y = ln
without simplifying it
2
first, we would have had:
 21
1
1
dy 4 x
1
2
y = ln 2 x

=
1
dx 1 x 2
2
=
1
1
2
1
2
2x x
1
=
2x
The derivative is the same, but the algebra is more difficult.
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The product rule
The chain rule
Contents
The relationship between
dy
dx
and
dx
dy
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
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The product rule
The product rule allows us to differentiate the product of two
functions.
It states that if y = uv, where u and v are functions of x, then
dy
dv
du
=u +v
dx
dx
dx
Find
dy
given that y = x 4 3  2 x .
dx
Let
u=
x4
du
= 4 x3
dx
So
and
v = (3  2 x)
1
2
dv 1
 21
= (3  2 x ) × 2
dx 2
 21
= (3  2 x)
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The product rule
Using the product rule:
1
dy
 21
4
3
2
=  x (3  2 x ) + 4 x (3  2 x)
dx
1
1
4
3
2
2
x
4 x (3  2 x ) (3  2 x )
=
+
1
1
2
2
(3  2 x )
(3  2 x )
=
=
=
 x 4 + 4 x3 (3  2 x )
(3  2 x )
1
2
 x 4 +12 x3  8 x 4
(3  2 x)
1
2
12 x3  9 x 4
(3  2 x )
1
2
3(4 x 3  3 x 4 )
=
3  2x
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The product rule
Give the coordinates of any stationary
points on the curve y = x2e2x.
u = x2
du
= 2x
dx
Let
So
and
v = e2 x
dv
= 2e2 x
dx
Using the product rule:
dy
= 2 x 2e2 x + 2 xe2 x
dx
= 2 xe2 x ( x +1)
dy
= 0 when 2 xe2 x = 0 or x +1= 0
dx
2 xe2 x = 0  x = 0
x +1= 0  x = 1
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The product rule
When x = 0, y = (0)2e0
=0
The point (0,0) is a stationary point on the curve y = x2e2x.
When x = –1, y = (–1)2e–2
= e–2
The point (–1, e–2) is also a stationary point on the curve
y = x2e2x.
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The quotient rule
The chain rule
Contents
The relationship between
dy
dx
and
dx
dy
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
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The quotient rule
The quotient rule allows us to differentiate the quotient of two
functions.
It states that if y = uv , where u and v are functions of x, then
dv

u
dy v du
= dx 2 dx
dx
v
dy
2 x +1
Find
given that y =
.
2
dx
5x
Let
u = 2x + 1
du
=2
dx
So
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and
v = 5x2
dv
= 10 x
dx
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The quotient rule
dy (5 x 2 )(2)  (2 x +1)(10 x)
=
dx
25 x 4
10 x 2  20 x 2  10 x
=
25 x 4
2x  4 x  2
=
5 x3
2 x  2
=
5 x3
2( x +1)
=
5 x3
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The quotient rule
Find the equation of the tangent to the
4
ln
x
curve y =
at the point (1, 0).
2
x
u = ln x4
and
v = x2
dv
du 4 x3
1
= 2x
= 4 = 4x
So
dx
dx
x
Using the quotient rule:
dy x 2 × 4 x 1  ln x 4 × 2 x
=
dx
x4
Let
4 x  8 x ln x
=
x4
4(1  2ln x )
=
x3
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Using ln x4 = 4 ln x
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The quotient rule
When x = 1, dy = 4(1  2 ln 1)
dx
1
Remember that ln 1 = 0
=4
 The gradient of the tangent at the point (1, 0) is 4.
Use y – y1 = m(x – x1) to find the equation of the tangent at the
point (1, 0).
y – 0 = 4(x – 1)
y = 4x – 4
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Differentiating trigonometric functions
The chain rule
Contents
The relationship between
dy
dx
and
dx
dy
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
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The derivative of sin x
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The derivative of sin x
By plotting the gradient function of y = sin x, where x is
measured in radians, we can deduce that
dy
If y = sin x then
= cos x
dx
Functions of the form k sin f(x) can be differentiated using the
chain rule.
Differentiate y = 2 sin 3x with respect to x.
So if
y = 2 sin u
dy
= 2cos u
du
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where
u = 3x
du
=3
dx
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The derivative of sin f(x)
Using the chain rule:
dy dy du
= 2cos u × 3
=
×
dx du dx
= 6cos3x
In general using the chain rule,
dy
If y = sin f ( x ) then
= f '( x)cos f ( x)
dx
Using
to represent f(x) and
y = sin
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to represent f ’(x):
dy

=
dy
cos
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The derivative of cos x
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The derivative of cos x
By plotting the gradient function of y = cos x, where x is
measured in radians, we can deduce that
dy
If y = cos x then
=  sin x
dx
dy
Find
given that y = –x2 cos x.
dx
Let
u = –x2
So
du
= 2 x
dx
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and
v = cos x
dv
=  sin x
dx
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The derivative of cos x
Using the product rule:
dy
=  x 2 (  sin x ) + cos x( 2 x)
dx
= x2 sin x  2x cos x
= x( x sin x  2cos x)
Functions of the form k cos f(x) can be differentiated using the
chain rule.
Differentiate y = 3 cos (x3 – 4) with respect to x.
So if
y = 3 cos u
dy
= 3 sin u
du
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where
u = x3 – 4
du
= 3 x2
dx
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The derivative of cos x
Using the chain rule:
dy dy du
=
×
= 3sin u ×3 x2
dx du dx
= 9 x 2 sin( x 3  4)
In general using the chain rule,
dy
If y = cos f ( x ) then
= f '( x)sin f ( x)
dx
Using
to represent f(x) and
y = cos
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
to represent f ’(x):
dy
=
dy
sin
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The derivative of tan x
We can differentiate y = tan x (where x is in radians) by writing it
as
sin x
y=
cos x
Then we apply the quotient rule with u = sin x and v = cos x :
dy cos x cos x  sin x(  sin x )
=
dx
cos2 x
cos2 x + sin2 x
=
cos2 x
1
=
cos2 x
= sec 2 x
If y = tan x then
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dy
= sec 2 x
dx
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The derivative of sec x
We can differentiate y = sec x (where x is in radians) by writing
it as
1
y=
= (cos x )1
cos x
Then using the chain rule we get:
dy
= (cos x )2 (  sin x)
dx
sin x
=
cos2 x
1
sin x
=
×
cos x cos x
= sec x tan x
If y = sec x then
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dy
= sec x tan x
dx
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The derivative of cosec x
We can differentiate y = cosec x (where x is in radians) by
writing it as
1
y=
= (sin x )1
sin x
Then using the chain rule we get:
dy
= (sin x )2 (cos x )
dx
cos x
= 2
sin x
1
cos x
=
×
sin x sin x
= cosec x cot x
If y = cosec x then
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dy
= cosec x cot x
dx
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The derivative of cot x
We can differentiate y = cot x (where x is in radians) by writing it
as
cos x
y=
sin x
Then we apply the quotient rule with u = cos x and v = sin x:
dy sin x(  sin x )  cos x cos x
=
dx
sin2 x
(sin2 x + cos2 x )
=
sin2 x
1
= 2
sin x
= cosec 2 x
If y = cot x then
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dy
=  cosec 2 x
dx
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Derivatives of trigonometric functions
In summary, if x is measured in radians, then
y = sin x 
dy
= cos x
dx
y = sec x 
dy
= sec x tan x
dx
dy
y = cos x 
=  sin x
dx
dy
y = cosec x 
= cosec x cot x
dx
dy
y = tan x 
= sec 2 x
dx
dy
y = cot x 
=  cosec 2 x
dx
When learning these results, it is helpful to notice that all of
the trigonometric functions starting with ‘co’ have negative
derivatives.
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Examination-style question
The chain rule
Contents
The relationship between
dy
dx
and
dx
dy
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
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Examination-style question
Given that f ( x ) =
2x
,
2
x +4
a) find f ’(x),
b) find the coordinates of any stationary points and determine
their nature,
c) sketch the curve y = f(x).
a) Using the quotient rule: f '( x ) =
dv
v du

u
dx
dx
v2
2( x 2 + 4)  2 x(2 x )
f '( x ) =
( x 2 + 4)2
2 x 2 + 8  4 x 2 2(4  x 2 )
=
= 2
2
2
( x + 4)
( x + 4)2
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Examination-style question
2(4  x 2 )
=0
b) When f(x) = 0,
2
2
( x + 4)
2(4  x 2 ) = 0
4  x2 = 0
x = 2
When x = 2, y =
4 1
=
8 2
4
1
=
8
2
2x
f(
x
)
=
Therefore, the graph of the function
has turning
2
x +4
points at (2, 21 ) and (–2, – 21).
When x = –2, y =
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Examination-style question
Looking at the gradient just before and just after x = 2:
x
dy 2(4  x 2 )
= 2
Value of
dx ( x + 4)2
Slope
1.9
2
2.1
0.01
0
–0.01
+ive
–ive
0
So (2, 21 ) is a maximum point.
Looking at the gradient just before and just after x = –2:
x
dy 2(4  x 2 )
= 2
Value of
dx ( x + 4)2
Slope
–2.1
–2
–1.9
0.01
0
–0.01
–ive
0
+ive
So (–2, – 21 ) is a minimum point.
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© Boardworks Ltd 2006
Examination-style question
2x
c) The curve y = 2
crosses the axes when x = 0 and when
x +4
y = 0.
When x = 0, y = 0.
(Also, when y = 0, x = 0).
Therefore the curve has one crossing point at the origin, a
minimum at (–2, – 21 ) and a maximum at (2, 21 ):
y
Also,
1
2
–2
0
 21
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y=
2
2x
x2 + 4
as x  , y  0–
x
and,
as x  , y  0+.
© Boardworks Ltd 2006