Download differentiation - Hinchingbrooke

DIFFERENTIATION
1. The Chain Rule (Function of a Function Rule)
Activity 1 : Expand the following functions, differentiate them, and then try and factorise the
derivative as much as possible. What do you notice?

y  (5 x  2)2

y  (3x  5)2

y  (2 x  3)3

y  ( x  3)4
Suppose we wish to differentiate y  (5 x  2)3 . One way is to expand the brackets and
differentiate each term individually, as in the activity above.
y  (5 x  2)3
 (5 x  2)(25 x 2  20 x  4)
 125 x 3  150 x 2  60 x  8
dy
 375 x 2  300 x  60
dx
 15(25 x 2  20 x  4)
 15(5 x  2) 2
This is lengthy method, and if the power of 5x  2 was much higher than 3, the process of
expanding the brackets would become very time-consuming.
A better way is to use the chain rule. This involves writing u  5x  2, y  u 3 . Notice that y is
now a function of u, which is itself a function of x. Hence y is a function of a function.
Now
dy
y
 lim
dx  x 0  x
 y u
 lim
 x 0  u  x
Now as  x  0 ,  u  0 , since u depends on x, and therefore the change in u depends on the
change in x. So we can write
dy
y
u
 lim
 lim

x

0

u

0
dx
u
x
dy du


du dx
This is the chain rule, sometimes called the function of a function rule.
dy dy du


dx du dx
We will now use it to differentiate our original function.
y  (5 x  2)3 , u  5x  2 ,
y  u3
dy dy du


dx du dx
 3u 2  5
 15(5 x  2) 2
This is simply the same answer as we got by expanding and then differentiating term-by-term.
Example 1 : Differentiate the following using the chain rule.
a) y   4 x 2  5

7
b) y  1  x

5
c) y 
1
x 3
2
d) y 
1
x3  1
Notice that the first two could be done by expanding and then differentiating term-by-term, but
the last one cannot. So the chain rule allows us to differentiate functions that were previously
impossible.
7
a)
c)
1
1
y   4 x 2  5 , u  4 x 2  5, y  u 7
y 2
, u  x 2  3, y 
x 3
u
dy dy du


dy dy du


dx du dx
dx du dx
6
 7u  8 x
1
7
  2  2x
2
 56 x  4 x  5 
u
2 x
5
b)

2
y  1  x , u  1  x , y  u5
2
 x  3
dy dy du
1
d)
1



3
3
y

,
u

x

1,
y

u
dx du dx
3 3
x 1
1
 5u 4 
dy dy du
2 x


dx du dx
4
4
5 1 x

  13 u 3  3 x 2

2 x
 x2

4
3
x

1
 3

3



Example 2 : Find the equations of the tangent and the normal to the curve y 
(1, –3).
y
3
, u  x2  2 ,
x 2
dy dy du


dx du dx
3
  2  2x
u
6 x

2
 x2  2
2
y
3
u
3
at the point
x 2
2
gradient of tangent 
6 1
1
2
 2
2
 6
gradient of normal 
1
6
The equation of the tangent is
y  y1  m( x  x1 )
y  3  6( x  1)
y  6 x  3
The equation of the normal is
y  y1  m( x  x1 )
y  3  16 ( x  1)
6 y  18  x  1
x  6 y  19
In general, if y   f ( x)  , i.e. a function raised to a certain power n, we have
n
y   f ( x)  ,
n
y  un
u  f ( x) ,
dy dy du


dx du dx
 nu n 1  f ( x)
 n f ( x)  f( x) 
n 1
To summarise…
y   f ( x) 
n

dy
n 1
 n f ( x) f ( x) 
dx
So the power drops by one, and both the original power and the derivative of the function inside
the brackets appear outside. Knowing this can save going through the chain rule every time - but
don’t take risks if you’re not sure!
Example 3 : Differentiate the following in your head.
a) y   5 x 2  7 x  1
3
a)
b)
2
dy
 3(10 x  7)  5 x 2  7 x  1
dx
y  3  x  2 x  11
3
2
3
dy
 3  2  3 x 2  2  x 3  2 x  11
dx
6  3 x 2  2 

3
 x3  2 x  11
b) y 
x
3
3
 2 x  11
c)
2
c) y 
y  (2 x  1)

1
2
3
dy

  12  2(2 x  1) 2
dx
1

3
(2 x  1) 2
1
2x 1
2. The Relationship Between
dy
dx
and
dx
dy
Consider the function y  x3 . Clearly
dy
 3x 2
dx
1
The function can be rearranged to make x the subject : x  y 3 . This gives us
dx 1  23
 y
dy 3
1

1
3 y3
 

Notice that
dy
dx
and
dx
dy
2
1
3x 2
are reciprocals of each other - as might be expected.
dx
1

dy  dy 
 
 dx 
2
Activity 2 : Verify that the above relationship holds with y  x 4 and y   .
x
This relationship gives us an alternative way of approaching problems that are seemingly
impossible...
Example 1 : Find the equation of the tangent to the curve xy  y 3  10 at the point (1, 2).
It is impossible to make y the subject of the formula and hence find
the subject of the formula, find
dx
dy
, and hence
dy
dx
.
10  y 3
y
10
  y2
y
dx
10
  2  2y
dy
y
x
At the point (1, 2),
dx
10
  2  2 2
dy
2
13

2
dy
2

dx
13
dy
dx
. However, we can make x
The equation of the tangent is of the form
y  y1  m( x  x1 )
2
( x  1)
13
13 y  26  2 x  2
y2 
2 x  13 y  28
C3 p122 Ex 8A
3. The Product Rule
Suppose we wish to differentiate y  3x 2 (4 x  1)3 . Rather than expanding this function first, we
use the product rule, since y can be written as the product of two functions of x, which we call u
and v.
So we have
y  3x 2 (4 x  1)3 , u  3 x 2 , v  (4 x  1)3 , y  uv
Let x, u, v and y change by small amounts δx, δu, δv and δy respectively. Then if y  uv ,
y   y  (u   u )(v   v)
 uv  u v  v u   u v
 y  u v  v u   u v
Dividing by δx,
y
v
u
v
u
v
u
x
x
x
x
Now as  x  0 ,
y
dy  u
du  v
dv



,
,
,  u  0 . This gives us the product rule...
x
dx  x
dx  x
dx
dy
dv
du
u v
dx
dx
dx
Going back to our original problem...
y  3x 2 (4 x  1)3 ,
u  3x 2 ,
v  (4 x  1)3 ,
y  uv
dy
dv
du
u v
dx
dx
dx
2
 3 x  12(4 x  1) 2  (4 x  1)3  6 x
 36 x 2 (4 x  1) 2  6 x(4 x  1)3
 6 x(4 x  1) 2  6 x  (4 x  1) 
 6 x(4 x  1) 2 (10 x  1)
Activity 2 : Check this result by first expanding the expression and differentiating as normal.
Example 1 : Differentiate the following.

a) y  x 4 1  x
a)

b) y  (3x  2)2 (5x  1)4
b)
dy
dv
du
u v
dx
dx
dx
1
 x4 
 1  x 4 x3
2 x

dy
dv
du
u v
dx
dx
dx
2
 (3x  2)  20(5 x  1)3  (5 x  1) 4  6(3 x  2)

 2(3x  2)(5 x  1)3 10(3 x  2)  3(5 x  1) 
 12 x3 x  4 x3  4 x3 x

 12 x3 9 x  8
 2(3x  2)(5 x  1)3 (45 x  23)

Example 2 : Find the stationary points on the curve y  (2 x  3)2 ( x  2) and determine their
natures. Hence sketch the curve.
dy
dv
du
u v
dx
dx
dx
2
 (2 x  3) 1  2  2(2 x  3)( x  2)
 (2 x  3)(2 x  3  4 x  8)
 (2 x  3)(6 x  5)


The stationary points are therefore  32 , 0 and

5
6

.
,  25 11
27
To determine their natures, we need the second derivative. We can do this by expanding
differentiating, or we can use the product rule again.
d2y
dv
du
u v
2
dx
dx
dx
 (2 x  3)  6  (6 x  5)  2
 24 x  8


At  32 , 0 ,
At

5
6
2
d y
 24   32  8  28 , so this is a maximum.
2
dx

,
,  25 11
27
d2y
 24  56  8  28 , so this is a minimum.
dx 2
y

3
2
,0

(2, 0)
x
(0,  18)

5
6
,  25 11
27

dy
and
dx
Example 3 : Find the equation of the normal to the curve y  x x  1 at the point (3, 6).
dy
dv
du
u v
dx
dx
dx

1
1
 x  12 ( x  1) 2  1 ( x  1) 2
x
 x 1
2 x 1
x  2( x  1)

2 x 1
3x  2

2 x 1

At (3, 6),
dy 3  3  2

dx 2 3  1
11

4
The equation of the normal is therefore
y  y1  m( x  x1 )
y  6   114 ( x  3)
11 y  66  4 x  12
4 x  11 y  78
C3 p123 Ex 8B
4. The Quotient Rule
We use the quotient rule when we want to differentiate functions of the form y 
u
,
v
Let x, u, v and y change by small amounts δx, δu, δv and δy respectively. Then if y 
have
u u
v v
u u u
y

v v v
v(u   u )  u (v   v)
y
v (v   v )
uv  v u  uv  u v

v (v   v )
v u  u v

v (v   v )
u
v
v
u
y
x
 x
x
v (v   v )
y  y 
u
we also
v
Now as  x  0 ,
y
dy  u
du  v
dv



,
,
,  v  0 . This gives us the quotient rule...
x
dx  x
dx  x
dx
dy

dx
Example 1 : Find
v
du
dv
u
dx
dx
2
v
dy
2x  3
if y 
,
1  5x
dx
y
2x  3
,
1  5x
u  2x  3 ,
v  1  5x ,
y
u
v
du
dv
u
dx
dx
2
v
(1  5 x)  2  5(2 x  3)

(1  5 x) 2
2  10 x  10 x  15

(1  5 x) 2
17

(1  5 x) 2
dy

dx
Example 2 : Find
v
dy
x2
if y 
dx
(4 x  3)3
y
x2
,
(4 x  3)3
u  x2 ,
v  (4 x  3)3 ,
y
u
v
du
dv
u
dx
dx
2
v
(4 x  3)3  2 x  x 2  12(4 x  3) 2

(4 x  3)6
dy

dx

v
x(4 x  3) 2  2(4 x  3)  12 x 
(4 x  3)6
x(6  4 x)

(4 x  3) 4
2 x(3  2 x)

(4 x  3) 4
C3 p125 Ex 8C
Traffic Flow and/or Knowing When to Stop
1
2
1
2
3
4
5
1– 2
2
1
5. Differentiation of Exponential Functions
We have already seen earlier in unit C3 that the function y  e x has the property that it remains
the same when differentiated.
dy
y  ex 
 ex
dx
y
5
y =e
x
4
3
2
1
–2
–1
1
2 x
Example 1 : Find the turning point of the function y  x  e x , and determine its nature.
y  x  ex
dy
 1  ex
dx
d2y
 e x
2
dx
At the turning point,
dy
0
dx
1  ex  0
ex  1
x0
So the turning point is at (0, 1). To determine its nature,
d2y
 e 0
2
dx
 1
So the turning point is a maximum.
Example 2 : Find the equation of the normal to the curve y  12 e x at the point where x  3 .
dy 1 x
 e
dx 2
At the point where x  3 ,
dy 1 3
 e
dx 2
The gradient of the normal is therefore 
2
. The normal passes through the point (3,
e3
1
2
e3 ) .
The equation of the normal is therefore
y  y1  m( x  x1 )
y  12 e3  
2
( x  3)
e3
We can, of course, use the chain, product and quotient rules with exponential functions...
Example 3 : Differentiate y  e x .
2
y  e x , u  x 2 , y  eu
dy dy du


dx du dx
 eu  2 x
2
 2 xe x
2
We can find a general rule for differentiating expressions of the form y  ef( x ) .
y  ef( x ) ,
u  f ( x) ,
y  eu
dy dy du


dx du dx
 eu  f ( x)
 f ( x)ef( x )
To summarise…
y  ef( x ) 
dy
 f ( x)ef( x )
dx
This enables us to differentiate ‘by inspection’.
Example 4 : Differentiate y  5e3 x
2
x
.
2
dy
 5(6 x  1)e3 x  x
dx
1
Example 5 : Differentiate y  6e x .
1
dy
1
 6  2  ex
dx
x
1
6 e x
 2
x
Example 6 : Differentiate y  x 2e x . Hence find the coordinates of the stationary points and
determine their natures.
y  x 2e x , u  x 2 , v  e x , y  uv
dy
dv
du
u v
dx
dx
dx
2 x
 x e  2 xe x
 x( x  2)e x
There are stationary points at x  0 and x  2 . The coordinates of the stationary points are
therefore (0, 0) and  2, 4e 2  .
We differentiate again to find the second derivative.
dy
 x( x  2)e x , u  x( x  2) ,
dx
v  ex ,
dy
 uv
dx
d2y
dv
du
u v
2
dx
dx
dx
 x( x  2)e x  (2 x  2)e x
  x2  4x  2 ex
3
2
1
4
5
6
7
8
1
2
3
3– 4
1
2
1
2
3
When x  0 ,
d2y
d2y
.
When
,
 2e 2 .

2
x


2
dx 2
dx 2
Therefore (0, 0) is a minimum, and  2, 4e 2  is a maximum.
We could also have reached this conclusion by sketching the graph of y  x 2e x from the graphs
of y  x 2 and y  e x …
y
y
4
8
7
2 x
6
y =x
5
3
2
y = xe
2
4
3
1
2
y=e
– 3
x
–2
1
– 1
1
2
3 x
–4
–3
–2
–1
1
2 x
x
.
e 4
x
y x
,
e 4
Example 7 : Differentiate y 
x
u  x,
v  ex  4 ,
y
u
v
du
dv
u
dy
 dx 2 dx
dx
v
x
(e  4)  1  xe x

(e x  4) 2
v

e x  4  xe x
(e x  4) 2
C3 p126 Ex 8D
6. Differentiation of Logarithmic Functions
Activity 3 : Get the class to work out the following integrals....
x
3
dx,  x 2 dx,  x1 dx,  x 0 dx,  x 1 dx,  x 2 dx,  x 3 dx .
They should notice that all the results are acceptable apart from
that we are about to resolve the mystery of integrating
x
dx . Explain
1
, something that was
x
surreptitiously avoided in unit C1!
A very important result is the differentiation of y  ln x .
Rewriting in index form,
x  ey
dx
 ey
dy
x
dy 1

dx x
And therefore
y  ln x 
1
dy 1

dx x
1
 x dx  ln x  c
This solves the problem we had in unit C1 when we were unable to integrate
1
.
x
Activity 4 : Plot the graph of y  ln x on Autograph. Then use slow plot and gradient function,
1
and check that the gradient function is the positive part of the y  graph. We will
x
solve the mystery of the negative part in unit C4!
Log laws make can make differentiating logarithmic functions a lot easier, as the next example
demonstrates.
Example 1 : Differentiate the following.
a) y  ln 5 x
b) y  ln x5
y  ln 5 x
a)
b)
y  ln x 5
 ln 5  ln x
 5ln x
dy
1
dy
1
 0

5

dx
x
dx
x
1
5


x
x
Example 2 : Find the equation of the tangent to the curve y  3ln x at the point where x  2 .
dy 3

dx x
At the point (2, 3ln 2) the gradient is
3
2
, so the equation of the tangent is
y  y1  m( x  x1 )
y  3ln 2  32 ( x  2)
y  32 x  3ln 2  3
Example 3 : Differentiate y  log10 x
We first change the base of the logarithm to e, using the standard base change formula.
y  log10 x

log e x
log e 10
1
 ln x
ln10
dy
1
1


dx ln10 x
1

x ln10

The following examples illustrate the use of the chain, product and quotient rules...
Example 4 : Differentiate y  ln  x 2  3 .
y  ln  x 2  3 ,
u  x2  3 ,
y  ln u
dy dy du


dx du dx
1
  2x
u
2x
 2
x 3
We can find a general rule for differentiating expressions of the form y  ln f( x) .
y  ln f( x) ,
u  f ( x) ,
y  ln u
dy dy du


dx du dx
1
  f ( x)
u
f ( x)

f( x)
To summarise…
y  ln f( x) 
dy f ( x)

dx f( x)
This enables us to differentiate ‘by inspection’.
Example 5 : Differentiate y  ln(e x  x) .
dy e x  1

dx e x  x
 x2  x 
Example 6 : Differentiate y  ln 

 2x  5 
Log laws are useful to ‘break up’ the function first.
 x2  x 
y  ln 

 2x  5 
 ln x  ln( x  1)  ln(2 x  5)
dy 1
1
2
 

dx x x  1 2 x  5
Example 7 : Differentiate y  x 2 ln x . Hence find
 x ln x dx .
u  x2 ,
y  x 2 ln x ,
v  ln x ,
y  uv
dy
dv
du
u v
dx
dx
dx
1
 x 2   2 x ln x
x
 x  2 x ln x
Therefore
 x  2 x ln x dx  x ln x  c
x  2  x ln x dx  x ln x  c
 x ln x dx  x ln x 
2
1
2
2
2
1
2
2
1
4
x2  c
Example 8 : Find the equation of the tangent to the curve y  e x ln x at the point (1, 0).
y  e x ln x ,
u  ex ,
v  ln x ,
dy
dv
du
u v
dx
dx
dx
1
 e x   e x ln x
x
x
e
 (1  x ln x)
x
At (1, 0)
dy e1
 (1  1ln1)
dx 1
e
The equation of the tangent is
y  y1  m( x  x1 )
y  0  e( x  1)
y  e( x  1)
y  uv
Example 9 : Differentiate y 
ln x
.
x
ln x
y
,
x
u  ln x ,
v  x,
y
u
v
du
dv
u
dy
 dx 2 dx
dx
v
1
x   ln x 1
x

x2
1  ln x

x2
v
C3 p128 Ex 8E
7. Differentiation of y  sin x and y  cos x
Activity 5 : Use Autograph to draw the graph of y  sin x , with the x-axis from 0 to 2π and the yaxis from –2 to 2. Make sure the angle mode is set to radians. Invite the class to
discuss what the gradient function will look like, and check by using the slow plot
and gradient function buttons. Repeat for y  cos x . The class should readily see
that…
dy
 cos x
dx
dy
y  cos x 
  sin x
dx
y  sin x 
y
y
1
1
y = sinx
0.5
0.5

– 0.5
–1
y = cosx
2
dy
= cosx
dx

3
2
2 x

– 0.5
–1
2

3
2
2 x
dy
= – sinx
dx
Activity 5 (continued) : Now set Autograph to degrees, and set the x-axis from 0 to 360. Repeat
the above activity and demonstrate the above rules no longer hold. Very able students
may wish to hypothesise what the gradient functions are this time, but there’s no
need, so long as it is appreciated why the gradient functions are ‘flattened’.

Example 1 : Find the equation of the normal to the curve y  sin x at the point  14  ,

dy
 cos x
dx
 cos 14 
1 
.
2
 

1
2
The gradient of the normal is therefore  2 .
y  y1  m( x  x1 )
y
1
  2( x  14  )
2
2 y  1  2 x  12 
2 x  2 y  12   1
8. Differentiation of Other Trigonometric Functions
Now we have the quotient rule, we are able to differentiate y  tan x .
sin x
, u  sin x , v  cos x ,
cos x
du
dv
v
u
dy
 dx 2 dx
dx
v
cos x  cos x  sin x   sin x

cos 2 x
sin 2 x  cos 2 x

cos 2 x
1

cos 2 x
 sec 2 x
dy
y  tan x 
 sec 2 x
dx
y  tan x 
1
5
4
3
2
1
2
3
4
1–
2
3
4
5
–5
1
2
3
4
5
3
2

22
y
u
v
y
5
4
3
2
1
–
–
–
–
–
1
2
3
4
5
2
dy
= sec x
dx


2
3
2
2 x
y = tanx
The graph shows that the gradient function does all the right things – in particular it is always
positive, since y  tan x is always an increasing function.
We can differentiate the remaining trigonometric functions as follows
y  sec x
y  cosec x
y  cot x
 (cos x) 1
u  cos x, y  u
 (sin x) 1
1
dy dy du

dx du dx
 u 2   sin x
u  sin x, y  u
cos x
sin x
u  cos x, v  sin x

1
dy dy du

dx du dx
 u 2  cos x
du
dv
u
dx
dx
v2
sin x   sin x  cos x   cos x

sin 2 x
  sin 2 x  cos 2 x 

cos 2 x
1
 2
sin x
  cosec 2 x
dy

dx
 cos x
sin 2 x
  cosec x cot x
sin x
cos 2 x
 sec x tan x


v
Activity 6 : Use Autograph to verify these results.
9. More Differentiation with Trigonometric Functions
Example 1 : Differentiate y  sin 2 x .
Using the chain rule,
u  2x ,
y  sin 2 x ,
y  sin u
dy dy du


dx du dx
 cos u  2
 2 cos 2 x
Activity 7 : By considering the graphs of y  sin x and y  sin 2 x , explain why the gradient
function in example 1 should be 2cos 2x and not just cos 2x . Again, you can use
Autograph to verify the result.
Example 2 : Differentiate y  tan  x 2  .
Using the chain rule,
y  tan  x 2  ,
u  x2 ,
dy dy du


dx du dx
 sec 2 u  2 x
 2 x sec 2  x 2 
y  tan u
Competent A-level students should be able to differentiate examples such as the ones above by
inspection. Consider the general case of y  sin f( x) .
y  sin f( x) ,
u  f( x) ,
y  sin u
dy dy du


dx du dx
 cos u  f ( x)
 f ( x) cos f( x)
This idea readily extends to the other trigonometric functions.
dy
y
dx
sin f( x)
f ( x) cos f( x)
cos f( x)
 f ( x) sin f( x)
tan f( x)
f ( x)sec2 f( x)
sec f( x)
f ( x) sec f( x) tan f( x)
cosec f( x)
 f ( x) cosec f( x) cot f( x)
cot f( x)
 f ( x) cosec2 f( x)
Example 3 : Differentiate y  tan  3x 2  5  .
dy
 6 x sec2  3 x 2  5 
dx
We now look at powers of trigonometric functions...
Example 4 : Differentiate y  sin 3 x .
Using the chain rule,
y  sin 3 x , u  sin x ,
y  u3
dy dy du


dx du dx
 3u 2  cos x
 3sin 2 x cos x
In general, for the function y  sin n x ,
y  sin n x ,
u  sin x ,
dy dy du


dx du dx
 nu n 1  cos x
 n sin n 1 x cos x
y  un
And for the other trigonometric functions...
y
dy
dx
sin n x
n sin n 1 x cos x
cos n x
n cos n 1 x sin x
tan n x
n tan n 1 x sec2 x
sec n x
n secn 1 x sec x tan x  n secn x tan x
cosecn x
n cosecn 1 x   cosec x cot x  n cosec n x cot x
cot n x
n cot n 1 x cosec2 x
Example 5 : Differentiate y  tan 5 x .
dy
 5 tan 4 x sec 2 x
dx
Example 6 : Differentiate y  cos6 x .
dy
 6 cos5 x sin x
dx
Example 7 : Differentiate y  sin 5  2 x3  .
Using a triple chain rule,
y  sin 5  2 x3  ,
v  2 x3 ,
u  sin v ,
y  u5
dy dy du dv

 
dx du dv dx
 5u 4  cos v  6 x 2
 30 x 2 cos  2 x 3  sin 4  2 x 3 
Finally, some examples using the product and quotient rules.
Example 8 : Differentiate y  x cos x . Hence find the equation of the normal to the curve at the
point

1
2
 , 0 .
Using the product rule,
y  x cos x ,
u  x,
v  cos x ,
dy
dv
du
u v
dx
dx
dx
 x   sin x  1 cos x
 cos x  x sin x
At the point

1
2
 , 0 .
 
dy
 cos 12   12  sin
dx
  12 
 
1
2
y  uv
The gradient of the normal is therefore
2

. The equation of the normal is of the form
y0 
y
2

2

( x  12  )
x 1
Example 9 : Show that the coordinates of the stationary points on the curve
y
sin x
x
satisfy the equation x  tan x .
Using the quotient rule,
du
dv
u
dy
 dx 2 dx
dx
v
cos x  x  1 sin x

x2
x cos x  sin x

x2
There are stationary points when this expression is equal to zero. Therefore
v
x cos x  sin x  0
x cos x  sin x
x  tan x
Activity 8 : Use Autograph to verify the answers to examples 1 and 2. For example 2, plot the
sin x
graphs of y 
, y  x and y  tan x . Check the last two intersect at the same xx
values as stationary point of the first.
C3 p130 Ex 8F, p131 Ex 8G, p132 Ex 8H, p133 Ex 8I, p134 Ex 8J
Topic Review : Differentiation