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:
n
If y = x th e n
dy
= nx
n 1
dx
and when xn is preceded by a constant multiplier k we have:
n
If y = kx th e n
dy
= knx
n 1
dx
Also:
If y = f ( x ) ± g ( x ) th e n
dy
= 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
e
3 x2
3
sin ( x )
Products of the form f(x) × g(x), such as:
x ln x
Quotients of the form
3x +1
x 1
2
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xe
f (x)
g( x )
2 x
2
3 x cos x
, such as:
e
2x
sin x
x
2
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:
3
3
2
(2 x + 1) = 8 x + 3( 4 x ) + 3(2 x ) + 1
3
2
= 8 x + 12 x + 6 x + 1
Differentiating with respect to x:
dy
2
= 24 x + 24 x + 6
dx
2
= 6( 4 x + 4 x + 1)
= 6(2 x + 1)
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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
=
dx
y = u3
So if
dy
= 3u
dy
×
du
du
dx
where
u = 2x + 1,
du
2
du
Using the chain rule:
=2
dx
dy
dx
=
dy
du
×
du
2
= 3u × 2
dx
= 6u
2
= 6(2 x + 1)
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2
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The chain rule
Use the chain rule to differentiate
y = 3 x 2  5 with respect to x.
Let
y=u
dy
du
=
1
2
1
2
u
u = 3x2 – 5
where
du
1
2
Using the chain rule,
= 6x
dx
dy
dx
=
dy
du
×
du
=
dx
1
2
u
1
2
= 3 xu
× 6x
1
2
= 3 x (3 x  5 )
2
=
1
2
3x
3x  5
2
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The chain rule
Find
dy
dx
Let
given that y =
y=
2
u
dy
4
= 8u
2
(7  x )
= 2u
3
4
4
.
du
5
du
Using the chain rule:
u = 7 – x3
where
= 3 x
2
dx
dy
dx
=
dy
du
×
du
= 8u
5
× 3 x
2
5
2
dx
= 24 x u
= 2 4 x (7  x )
2
=
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24 x
3
5
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)
dy
then
where
and
= g '( u )
du
Using the chain rule:
dy
dx
=
dy
×
du
du
u = f(x)
du
= f '( x )
dx
= g '( u ) × f '( x )
dx
But u = f(x) so
If y = g(f(x)) then
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dy
= g '( f ( x )) × f '( x )
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
then
dy
= n ( f ( x ))
y=

× f '( x )
dx
If we use
to represent f (x) and
write this rule more visually as:
n
n 1
dy
=n
to represent f ’(x) we can
n 1
dy
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
= 3( x  3 ) ×
4
2
dx
4
2
= 12 x ( x  3)
3
When x = 1,
4
dx
= 3( x  3 ) × 4 x
dy
( x  3)
4
3
2
= 1 2(1  3 ) = 4 8
2
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 d y by differentiating with respect to y:
dx
= 8y
dy
Using the chain rule we can write
dy
×
dx
dx
dy
= 1,
from which we get:
So by the above result, if
dx
=
dx
=
1
dx
dy
8y then
dy
dy
dx
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dy
=
1
8y
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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
= 6y 3
2
dy
At the point (3, 2), y = 2:
dx
= 6 (2 )  3 = 2 1
2
dy
We can now find the gradient using the fact that
dy
dx
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=
dy
dx
=
1
dx
dy
1
21
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Differentiating inverse functions
The result
dy
=
dx
1
dx
dy
is particularly useful for differentiating
inverse functions. For example:
Find
d
(sin
dy
1
x ), writing your answer in terms of x.
Let y = sin–1 x so
x = sin y
dx
Using the identity
cos2y = 1 – sin2y
= co s y
dy
dy
=
dx
1
=
1
1  s in y
2
co s y
But sin y = x so
d
dy
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(sin
1
x) =
1
1 x
2
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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
x
If y = e th e n
=e
x
dx
From this, it follows that
dy
x
If y = ke th e n
= ke
x
dx
where k is a constant.
For example,
if y = 4ex – x3
dy
= 4e  3 x
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.
Let
y =e
dy
=e
u
where
u = 5x
du
u
dx
du
Using the chain rule:
=5
dy
dx
=
dy
du
×
du
dx
u
= e ×5
= 5e
u
= 5e
5x
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
th e n
dy
= ke
kx
dx
For example,
d
dx
d
(e
(e
7x
) = 7e
2 x
7x
) = 2e
2 x
dx
d
dx
x
3
(e ) =
e
x
3
3
We can use the chain rule to extend this to any function of the
form ef(x).
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Functions of the form ef(x)
If y = ef(x) then we can let
Let
y =e
dy
then
=e
u
where
u = f(x)
du
u
du
= f '( x )
dx
Using the chain rule:
dy
=
dx
dy
×
du
du
dx
u
= e × f '( x )
= f '( x ) e
So in general,
If y = e
f (x)
th e n
dy
= f '( x ) e
f (x)
f (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):
to represent f(x) and

y=e
dy
=
e
dx
For example,
dy
(e
5x4
) = 5e
5x4
dx
dy
(e
3x
) = e
3x
dx
dy
2
2
(5 e
x 9
) = 5 × 2 xe
x 9
2
= 10 xe
x 9
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
=e
y
dy
dy
But
ey
= x so,
=
1
dx
dx
dy
dy
1
dx
=
=
1
e
y
x
If y = ln x th e n
dy
dx
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=
1
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
where
y = ln u
dy
du
=
u = 3x
1
du
u
dx
Using the chain rule:
dy
dx
=
dy
du
×
du
=
=3
3
u
dx
=
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,
If y = ln kx th e n
dy
=
1
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)
dy
then
du
=
1
du
u
dx
Using the chain rule:
dy
dx
=
dy
du
×
du
=
dx
= f '( x )
f '( x )
u
=
f '( x )
f (x)
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Functions of the form ln (f(x))
In general, using the chain rule
If y = ln f ( x ) th e n
dy
=
f '( x )
f (x)
dx
Using
to represent f(x) and
y = ln

to represent f ’(x):
dy
=
dy
For example,
d
dx
7x  4
d
9x
dx
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ln (7 x  4 ) =
7
3
ln (3 x + 8 ) =
2
3
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
Differentiate y = ln
y = ln
x
2
= ln
ln
a
= ln a  ln b
b
x
2
ln an = n ln a
with respect to x.
x  ln 2
1
2
= ln x  ln 2
=
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1
2
ln x  ln 2
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Functions of the form ln (f(x))
y=
1
2
ln x  ln 2
dy

dx
=
1
2
×
1
x
1
=
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:

y = ln
1
2
x
1
2

dy
=
1
4
1
2
dx
=
=
1
2
x
x
1
2
1
1
2
2x x
1
1
2
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
dx
F in d
dy
g iv e n th a t y = x
4
=u
dv
+v
dx
du
dx
3  2 x.
dx
Let
u=
du
So
dx
x4
= 4x
and
3
v = (3  2 x )
dv
dx
=
1
2
1
2
(3  2 x )
=  (3  2 x )
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 21
× 2
 21
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The product rule
Using the product rule:
dy
=  x (3  2 x )
4
dx
=
x
+
1
2
(3  2 x )
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1
2
1
2
 x + 12 x  8 x
3
(3  2 x )
12 x  9 x
(3  2 x )
4
1
2
4
1
2
3( 4 x  3 x )
3
=
(3  2 x )
1
2
3
3
=
1
2
 x + 4 x (3  2 x )
4
=
1
2
3
4
=
+ 4 x (3  2 x )
3
4 x (3  2 x ) (3  2 x )
4
(3  2 x )
 21
3  2x
4
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The product rule
Give the coordinates of any stationary
points on the curve y = x2e2x.
u = x2
Let
du
So
v = e2x
and
dv
= 2x
dx
= 2e
2x
dx
Using the product rule:
dy
2
= 2x e
2x
+ 2 xe
2x
dx
2x
= 2 xe ( x + 1)
dy
= 0 w h e n 2 xe
2x
= 0 or x + 1 = 0
dx
2 xe
2x
=0 
x +1= 0 
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x=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
dy
=
v
dx
Find
dy
dx
Let
given that y =
2x +1
u = 2x + 1
du
So
dx
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=2
5x
2
du
dx
u
v
dv
dx
2
.
and
v = 5x2
dv
= 10 x
dx
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The quotient rule
dy
(5 x )(2 )  (2 x + 1)(1 0 x )
2
=
25 x
dx
10 x  20 x  10 x
2
=
2
25 x
=
3
2 x  2
5x
=
3
2( x + 1)
5x
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4
2x  4x  2
5x
=
4
3
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The quotient rule
Find the equation of the tangent to the
curve y =
ln x
x
4
at the point (1, 0).
2
u = ln x4
Let
du
So
dx
=
3
4x
x
v = x2
and
4
= 4x
dv
1
= 2x
dx
Using the quotient rule:
dy
2
=
x × 4x
dx
 ln x × 2 x
4
x
=
=
4
4 x  8 x ln x
x
Using ln x4 = 4 ln x
4
4 (1  2 ln x )
x
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1
3
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The quotient rule
When x = 1, dy = 4 (1  2 ln 1)
1
dx
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
39 of 56
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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
If y = sin x th e n
dy
= co s 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
du
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= 2 co s u
where
u = 3x
du
=3
dx
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The derivative of sin f(x)
Using the chain rule:
dy
dx
=
dy
du
×
du
= 2 cos u × 3
dx
= 6 co s 3 x
In general using the chain rule,
If y = sin f ( x ) th e n
dy
= f '( x ) co s f ( x )
dx
Using
to represent f ’(x):
to represent f(x) and
y = sin

dy
=
co s
dy
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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
If y = co s x th e n
dy
=  sin x
dx
Find
dy
dx
given that y = –x2 cos x.
u = –x2
Let
du
So
dx
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= 2 x
and
v = cos x
dv
=  sin x
dx
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The derivative of cos x
Using the product rule:
dy
=  x (  sin x ) + co s x (  2 x )
2
dx
= x sin x  2 x cos x
2
= x ( x sin x  2 cos 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
du
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=  3 sin u
where
u = x3 – 4
du
= 3x
2
dx
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The derivative of cos x
Using the chain rule:
dy
dx
=
dy
×
du
du
=  3 sin u × 3 x
2
dx
=  9 x sin ( x  4 )
2
3
In general using the chain rule,
If y = co s f ( x ) th e n
dy
=  f '( x ) sin f ( x )
dx
Using
to represent f(x) and
y = co s

dy
to represent f ’(x):
=
sin
dy
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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=
co s x
Then we apply the quotient rule with u = sin x and v = cos x :
dy
=
co s x co s x  sin x (  sin x )
2
co s x
dx
2
=
2
co s x + sin x
2
co s x
=
1
2
co s x
2
= sec x
If y = ta n x th e n
dy
2
= se c 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
1
y=
= (co s x )
co s x
Then using the chain rule we get:
dy
2
=  (co s x ) (  sin x )
dx
=
sin x
2
co s x
=
1
×
co s x
sin x
co s x
= se c x ta n x
If y = se c x th e n
dy
= se c x ta n 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
1
y=
= (sin x )
sin x
Then using the chain rule we get:
dy
2
=  (sin x ) (co s x )
dx
=
co s x
2
sin x
=
1
×
sin x
co s x
sin x
=  cosec x cot x
If y = co se c x th e n
dy
=  co se c x co t 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
co s x
y=
sin x
Then we apply the quotient rule with u = cos x and v = sin x:
dy
=
sin x (  sin x )  co s x co s x
2
sin x
dx
 (sin x + co s x )
2
=
2
2
sin x
=
1
2
sin x
=  co sec x
2
If y = co t x th e n
dy
=  co se c x
2
dx
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Derivatives of trigonometric functions
In summary, if x is measured in radians, then
y = sin x 
dy
= co s x
y = se c x 
dx
y = co s x 
dy
y = ta n x 
dx
= se c x ta n x
dx
=  sin x
y = co se c x 
dx
dy
dy
dy
=  co se c x co t x
dx
2
= se c x
y = co t x 
dy
=  co se c x
2
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).
v
a) Using the quotient rule: f '( x ) =
du
dx
u
v
dv
dx
2
2( x + 4 )  2 x (2 x )
2
f '( x ) =
( x + 4)
2
2
2x + 8  4x
2
2
=
2
(x + 4)
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2
2 (4  x )
2
=
2
(x + 4)
2
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Examination-style question
2 (4  x )
2
b) When f(x) = 0,
2
(x + 4)
2
=0
2 (4  x ) = 0
2
4x =0
2
x = 2
When x = 2, y =
4
=
8
When x = –2, y =
1
2
4
8
=
1
2
Therefore, the graph of the function f( x ) =
points at (2, 21 ) and (–2, – 21 ).
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2x
2
x +4
has turning
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Examination-style question
Looking at the gradient just before and just after x = 2:
x
Value of
dy
2 (4  x )
1.9
2
2.1
0.01
0
–0.01
2
=
2
(x + 4)
dx
2
+ive
Slope
–ive
0
So (2, 21 ) is a maximum point.
Looking at the gradient just before and just after x = –2:
x
Value of
dy
dx
2 (4  x )
–2.1
–2
–1.9
0.01
0
–0.01
2
=
2
(x + 4)
Slope
2
–ive
0
+ive
So (–2, – 21 ) is a minimum point.
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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

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1
2
y=
2
2x
as x   , y  0–
2
x +4
x
and,
as x   , y  0+.
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