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A-Level Maths: Core 3for Edexcel
C3.6 Differentiation
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The chain rule
The relationship between
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
dy dxdx dyand
Co
nte
nts
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The chain rule
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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:
and when xn is preceded by a constant multiplier k we have:
1If = then =n ndyy kx knx
dx
f g f gIf = ( ) ± ( ) then = '( ) ± '( )dy
y x x x xdx
Also:
1If = then =n ndyy x nx
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:
2 1x xe 3 2 x3sin( )
Products of the form f(x) × g(x), such as:
x xln xxe 2 x x23 cos
x
x 2
3 +1
1
xe
x
2
sin
x
x
2
ln
Quotients of the form , such as:f
g
( )
( )
x
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:
x x x x3 3 2(2 +1) = 8 + 3(4 )+ 3(2 )+1
x x x3 2= 8 +12 + 6 +1
dyx x
dx2= 24 + 24 + 6
x x2= 6(4 + 4 +1)
x 2= 6(2 +1)
Differentiating with respect to x:
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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, thendy dy du
dx du dx= ×
So if y = u3 where u = 2x + 1,dy
udu
2= 3du
dx= 2
Using the chain rule:dy dy du
dx du dx= × u2= 3 ×2
u2= 6
x 2= 6(2 +1)
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The chain rule
121
2=dy
udu
dux
dx= 6
Using the chain rule,dy dy du
dx du dx= ×
121
2= ×6u x
122= 3 (3 5)x x
Use the chain rule to differentiate y = with respect to x.x 23 5
Let where u = 3x2 – 5y u12=
2
3=
3 5
x
x
xu 12= 3
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The chain rule
dyu
du 5= 8
dux
dx 2= 3
Using the chain rule:dy dy du
dx du dx= × u x 5 2= 8 × 3
x x 2 3 5= 24 (7 )
Let yu4
2=
x
x
2
3 5
24=
(7 )
x u 2 5= 24
Find given that .yx 3 4
2=
(7 )
dy
dx
u 4= 2 where u = 7 – x3
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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)
then andg= '( )dy
udu
f= '( )du
xdx
Using the chain rule:dy dy du
dx du dx= × g f= '( )× '( )u x
But u = f(x) so
If y = g(f(x)) then g f f= '( ( ))× '( )dy
x xdx
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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 f f1= ( ( )) × '( )ndyn x x
dx
If we use to represent f (x) and to represent f ’(x) we can write this rule more visually as:
1= = n ndyy n
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
When x = 1,dy
dx 2=12(1 3) = 48
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)
dy dx x
dx dx 4 2 4= 3( 3) × ( 3)
x x4 2 3= 3( 3) × 4
x x 3 4 2=12 ( 3)
y = (x4 – 3)3
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The chain rule
The relationship between
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
dy dxdx dyand
Co
nte
nts
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The relationship between dy dxdx dyand
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The relationship between
Suppose we are given x as a function of y instead of y as a function of x. For instance,
dy dxdx dyand
x = 4y2
We can find by differentiating with respect to y:dxdy
= 8dx
ydy
Using the chain rule we can write
× =1,dy dx
dx dy
1=
dxdy
dy
dx
So by the above result, if 8y then=dx
dy1
=8
dy
dx y
from which we get:
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The relationship between
Find the gradient of the curve with equation x = 2y3 – 3y – 7 at the point (3, 2).
dy dxdx dyand
x = 2y3 – 3y – 7
dxy
dy2= 6 3
At the point (3, 2), y = 2:
dx
dy2= 6(2) 3 = 21
We can now find the gradient using the fact that 1
=dxdy
dy
dx
dy
dx
1=
21
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Differentiating inverse functions
= cosdx
ydy
Find , writing your answer in terms of x.1(sin )
dx
dy
Let y = sin–1 x so
The result is particularly useful for differentiating
inverse functions. For example:
1=
dxdy
dy
dx
x = sin y
Using the identity cos2y = 1 – sin2y
But sin y = x so
1=
cos
dy
dx y 2
1=
1 sin y
1
2
1(sin ) =
1
dx
dy x
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Co
nte
nts
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Differentiating ex and related functions
The chain rule
The relationship between
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
dy dxdx dyand
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The derivative of ex
From this, it follows that
x xdyy e e
dxIf = then =
x xdyy ke ke
dxIf = then =
A special property of the exponential function ex is that
where k is a constant.
For example, if y = 4ex – x3
2= 4 3xdye x
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.
udye
du= = 5
du
dx
Using the chain rule:dy dy du
dx du dx= × = ×5ue
5= 5 xe
Let where u = 5xuy e=
= 5 ue
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,
For example,
We can use the chain rule to extend this to any function of the form ef(x).
If = then =kx kxdyy e ke
dx
2 2( ) = 2x xde e
dx
7 7( ) = 7x xde e
dx
33( ) =
3
xxd e
edx
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Functions of the form ef(x)
If y = ef(x) then we can let
Using the chain rule:dy dy du
dx du dx= × f= × '( )ue x
Let where u = f(x)uy e=
ff ( )= '( ) xx e
f= '( )du
xdx
udye
du=then
So in general, f ff( ) ( )If = then = '( )x xdyy e x e
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)
For example,
Using to represent f(x) and to represent f ’(x):
= = dy
y e edx
3( ) =xdye
dx 3 xe
5 4( ) =xdye
dx 5 45 xe
2 9(5 ) =xdye
dx 2 95×2 =xxe 2 910 xxe
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The chain rule
The relationship between
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
dy dxdx dyand
Co
nte
nts
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Differentiating ln x and related functions
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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:
ydxe
dy=
1 1= =
ydxdy
dy
dx e
dy
dx x
1=But ey = x so,
dyy x
dx x
1If = ln then =
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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.
1=
dy
du u= 3
du
dx
Using the chain rule:dy dy du
dx du dx= ×
3=
u
1=
x
Let where u = 3x= lny u
3=
3x
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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,
We can use the chain rule to extend to functions of the more general form y = ln f(x).
1If = ln then =
dyy kx
dx x
Using the chain rule:dy dy du
dx du dx= ×
f '( )=
x
u
Let where u = f(x)= lny u
f
f
'( )=
( )
x
x
f= '( )du
xdx
1=
dy
du uthen
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Functions of the form ln (f(x))
ff
f
'( )If = ln ( ) then =
( )
dy xy x
dx x
In general, using the chain rule
Using to represent f(x) and to represent f ’(x):
For example, ln(7 4) =d
xdx
7
7 4x
= ln =
dyy
dy
3ln(3 + 8) =d
xdx
2
3
9
3 + 8
x
x
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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 ln (ab) = ln a + ln b ln = ln lna
a bb
ln an = n ln a ln an = n ln a
Differentiate with respect to x.= ln2
xy
= ln2
xy = ln ln2x
12= ln ln2x
12= ln ln2x
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Functions of the form ln (f(x))
ln 2 is a constant and so it disappears when we differentiate.
12= ln ln2 =
dyy x
dx
1 1= ×
2 x
1=
2x
If we had tried to differentiate without simplifying it first, we would have had:
= ln2
xy
121
2= ln =dy
y xdx
12
12
1412
x
x
1 12 2
1=
2x x1
=2x
The derivative is the same, but the algebra is more difficult.
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The chain rule
The relationship between
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
dy dxdx dyand
Co
nte
nts
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The product rule
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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 duu v
dx dx dx= +
So3= 4
dux
dx
12
1= (3 2 ) × 2
2
dvx
dx
4Find given that = 3 2 .dy
y x xdx
Let u = x4 and v 12= (3 2 )x
12= (3 2 )x
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The product rule
Using the product rule:1 12 24 3= (3 2 ) + 4 (3 2 )
dyx x x x
dx
1 12 2
1 12 2
4 34 (3 2 ) (3 2 )= +
(3 2 ) (3 2 )
x x x x
x x
12
4 3+ 4 (3 2 )=
(3 2 )
x x x
x
12
4 3 4+12 8=
(3 2 )
x x x
x
12
3 412 9=
(3 2 )
x x
x
3 43(4 3 )
=3 2
x x
x
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The product rule
Give the coordinates of any stationary points on the curve y = x2e2x.
So = 2du
xdx
2= 2 xdve
dx
Let u = x2 and v 2= xe
Using the product rule:
2 2 2= 2 + 2x xdyx e xe
dx2= 2 ( +1)xxe x
2= 0 when 2 = 0 or +1= 0xdyxe x
dx22 = 0 = 0xxe x
+1= 0 = 1x x
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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 chain rule
The relationship between
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
dy dxdx dyand
Co
nte
nts
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The quotient rule
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The quotient rule
The quotient rule allows us to differentiate the quotient of two functions.
2=
du dvdx dxv udy
dx v
It states that if y = , where u and v are functions of x, thenuv
Find given that .x
yx2
2 +1=
5
dy
dx
Let u = 2x + 1 and v = 5x2
Sodu
dx= 2
dvx
dx=10
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The quotient rule
dy x x x
dx x
2
4
(5 )(2) (2 +1)(10 )=
25
x x x
x
2 2
4
10 20 10=
25
x x
x
3
2 4 2=
5
x
x
3
2 2=
5
x
x
3
2( +1)=
5
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The quotient rule
Let u = ln x4 and v = x2
So3
4
4=
du x
dx x= 2
dvx
dx
Using the quotient rule:2 1 4
4
× 4 ln ×2=
dy x x x x
dx x
4
4 8 ln=
x x x
x
Find the equation of the tangent to the
curve y = at the point (1, 0).4
2
ln x
x
1= 4x
3
4(1 2ln )=
x
x
Using ln x4 = 4 ln x
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The quotient rule
When x = 1,
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
4(1 2 ln 1)=
1
dy
dx
= 4
Remember that ln 1 = 0
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The chain rule
The relationship between
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
dy dxdx dyand
Co
nte
nts
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Differentiating trigonometric functions
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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
dyy x x
dxIf = sin then = cos
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 where u = 3x
= 2cosdy
udu
= 3du
dx
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The derivative of sin f(x)
Using to represent f(x) and to represent f ’(x):
= sin = cosdy
ydy
Using the chain rule:dy dy du
dx du dx= × = 2cos ×3u
= 6cos3x
In general using the chain rule,
f f fIf = sin ( ) then = '( )cos ( )dy
y x x xdx
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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
dyy x x
dxIf = cos then = sin
Find given that y = –x2 cos x.dy
dx
Let u = –x2 and v = cos x
So = 2du
xdx
= sindv
xdx
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The derivative of cos x
Using the product rule:
2= ( sin )+ cos ( 2 )dy
x x x xdx
2= sin 2 cosx x x x
= ( sin 2cos )x x x 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 where u = x3 – 4
= 3sindy
udu
2= 3du
xdx
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The derivative of cos x
Using to represent f(x) and to represent f ’(x):
= cos = sindy
ydy
Using the chain rule:dy dy du
dx du dx= × 2= 3sin ×3u x
2 3= 9 sin( 4)x x
In general using the chain rule,
f f fIf = cos ( ) then = '( )sin ( )dy
y x x xdx
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The derivative of tan x
We can differentiate y = tan x (where x is in radians) by writing it as x
yx
sin=
cosThen we apply the quotient rule with u = sin x and v = cos x :
dy x x x x
dx x
2
cos cos sin ( sin )=
cos
x x
x
2 2
2
cos + sin=
cos
x2
1=
cos
x2= sec
dyy x x
dx2If = tan then = sec
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The derivative of sec x
We can differentiate y = sec x (where x is in radians) by writing it as
y xx
11= = (cos )
cos
Then using the chain rule we get:dy
x xdx
2= (cos ) ( sin )
x
x2
sin=
cos
x
x x
1 sin= ×
cos cos
x x= sec tan
dyy x x x
dxIf = sec then = sec tan
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The derivative of cosec x
We can differentiate y = cosec x (where x is in radians) by writing it as
y xx
11= = (sin )
sin
Then using the chain rule we get:dy
x xdx
2= (sin ) (cos )
x
x
2
cos=
sin
x
x x
1 cos= ×
sin sinx x= cosec cot
dyy x x x
dxIf = cosec then = cosec cot
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The derivative of cot x
We can differentiate y = cot x (where x is in radians) by writing it as x
yx
cos=
sinThen we apply the quotient rule with u = cos x and v = sin x:
dy x x x x
dx x
2
sin ( sin ) cos cos=
sin
x x
x
2 2
2
(sin + cos )=
sin
x
2
1=
sin
x 2= cosec
2If = cot then = cosec dy
y x xdx
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dyy x x
dx= sin = cos
Derivatives of trigonometric functions
In summary, if x is measured in radians, then
2= cot = cosec dy
y x xdx
dyy x x x
dx= sec = sec tan
dyy x x x
dx = cosec = cosec cot
dyy x x
dx = cos = sin
dyy x x
dx 2= tan = sec
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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The chain rule
The relationship between
Differentiating ex and related functions
Differentiating ln x and related functions
The product rule
The quotient rule
Differentiating trigonometric functions
Examination-style question
dy dxdx dyand
Co
nte
nts
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Examination-style question
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Examination-style question
a) find f ’(x),
b) find the coordinates of any stationary points and determine their nature,
c) sketch the curve y = f(x).
Given that , f2
2( ) =
+ 4
xx
x
a) Using the quotient rule: f 2'( ) =
du dvdx dxv u
xv
f2
2 2
2( + 4) 2 (2 )'( ) =
( + 4)
x x xx
x
x x
x
2 2
2 2
2 + 8 4=
( + 4)
2
2 2
2(4 )=
( + 4)
x
x
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Examination-style question
b) When f(x) = 0,x
x
2
2 2
2(4 )= 0
( + 4)
x 22(4 ) = 0
x 24 = 0
x = 2
When x = 2, y4
= 8
1=
2
When x = –2, y 4
= 8
1
= 2
Therefore, the graph of the function has turning points at (2, ) and (–2, – ).
xx
x2
2f( ) =
+ 412
12
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Examination-style question
Looking at the gradient just before and just after x = 2:
12So (2, ) is a maximum point.
x 1.9 2 2.1
Value of2
2 2
2(4 )=
( + 4)
dy x
dx x
Slope
Looking at the gradient just before and just after x = –2:
12So (–2, – ) is a minimum point.
x –2.1 –2 –1.9
Value of2
2 2
2(4 )=
( + 4)
dy x
dx x
Slope
–ive0
+ive
0.01 0 –0.01
0.01 0 –0.01
–ive 0 +ive
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Examination-style question
When x = 0,
c) The curve crosses the axes when x = 0 and when y = 0.
2
2=
+ 4
xy
x
y = 0. (Also, when y = 0, x = 0).
Therefore the curve has one crossing point at the origin, a minimum at (–2, – ) and a maximum at (2, ):1
212
Also,
as , x y 0–
0 and,
as , x y 0+.
0
2
2=
+ 4
xy
x
0–2 212
12
x
y