More about differentiation.
Verb: To differentiate means to find the rate of growth (or
decay) of something.
Noun: Differential (or the derivative) – The rate of growth
Noun: Differentiation – The process that finds the rate of
growth.
Adjective: Differentiable – Qualifier for the entity whose
growth rate is measurable in some way. A function that can
be differentiated
A differentiable function is the one for which derivative
could be computed.
For that to be possible, we need (at the computational
point)
■ The function to be continuous (without break, etc.)
■ The function to be smooth (without sharp point)
■ The function to have a definite slope
Slope of a function y  f ( x )at a point ( x , y ) is its
differential computed at that point, i.e.
slope  m 
dy
dx ( x , y )
Example. y  f ( x )  3 x 2  7 x  3 . Find the slope of the
tangent at the point (2, 1).
dy
 6 x 7
dx
At (2,1), m  5 . The equation of the line with a slope of 5
passing through (2,1) point is:
Slope of f ( x ) is: m 
y  1  5( x  2 ) or y  5 x  9
This is the tangent line on the function at the point (2,1).
We can compute differentials directly on an equation.
e.g. Let
y 2  x3  3 x 2  6 x  12
differentiating both sides with respect to x we get,
2y
dy
 3 x2  6 x  6 .
dx
dy 3 x 2  6 x  6
3 x2  6 x  6


Therefore,
dx
2y
2 x 3  3 x 2  6 x  12
Another example.
sin y  x3  4 x 2  2
Differentiating both sides with respect to x, we get
dy
 3 x2  8 x
dx
dy 3 x 2  8 x

Therefore,
dx
cos y
cos y
Differentiation of exponential and logarithmic functions.
1. Let y  eax , where a is a constant.
dy
e a( x  h )  e ax
 lim
Then
dx h 0
h
Now, ea( x  h )  eax  eax .eah  eax  eax ( eah  1 )
a 2h2
 ...  1  ah for small h.
However, e  1  1  ah 
2!
dy
e a( x  h )  e ax ahe ax
 lim

 ae ax
Therefore,
dx h 0
h
h
ah
2. Let y  ln x
Then
dy
ln( x  h )  ln( x )
 lim
dx h0
h
h
 x  h

Now, ln( x  h )  ln( x )  ln
  ln 1  
x
 h 

h
h
If h  0 , ln( 1  ) 
x
x
Therefore,
dy
ln( x  h )  ln( x ) h 1
 lim


dx h0
h
xh x
3. y  a x . Find
dy
dx
Take log both sides of y  a x . Then we get
ln y  x ln a .
If we differentiate it with respect to x , we get
1 dy
dy
 ln a , or
 y ln a
y dx
dx
Product rule.
If y  f ( x )  u( x )v( x ) or simply y  uv , then
dy
dv
du
 u  v
dx
dx
dx
Example. 1. y  ( x 2  3 )( x4  2 x  3 )
Then
dy
 ( x 2  3 )  ( 4 x3  2 )  ( x4  2 x  3 )  2 x
dx
2. y  x sin x
dy
 x cos x  sin x
dx
3. y  x  2 ( 4  3 x 3 )
dy
d
d
 x  2 ( 4  3 x 3 )  ( 4  3 x 3 ) x  2
dx
dx
dx
Now,
and,
d
( 4  3 x 3 )  3  ( 3 )  x 4  9 x 4
dx
d 2
( x )  2 x 3
dx
Substituting these in our differential expression, we get
dy
 9 x  2 x 4  2( 4  3 x 3 )x 3
dx
 9 x 6  8 x 3  6 x 6  ( 15 x 6  8 x 3 )
Exercises for homework.
Find differentials of the following expressions.
a. y  x 3 ln x
b. y  3 x4  sin x cos x
c. y  e  x ( 3  x )
How do we compute the differential for a function which is
a product of three functions? Four functions? n functions?
Suppose, y  uvw where all of them are functions of x .
Then
But
dy
d
d
 u vw  vw u
dx
dx
dx
d
dw
dv
vw  v
w
dx
dx
dx
Plugging this above, we get
dy
 u( vw  wv )  vwu  uvw  uvw  uvw
dx
Similarly, for product of four functions, etc.
Example. Let y  x  2 ( 3  x3 )( 1  x )
d 2
d
x  2 x 3 ,
( 3  x3 )  3 x 2 and
dx
dx
d
1
1
( 1  x )  x 1 / 2 
dx
2
2 x
Note that
Therefore,
dy
 2 x 3 ( 3  x3 )( 1  x )  3 x 2 x  2 ( 1  x )
dx
x 2 ( 3  x3 )
+
2 x
Division rule.
Suppose y ( x) 
u
u ( x)
i.e. y 
v( x)
v
Then,
dy v  u( x)  u  v( x)

dx
v2
Example.
x2
dy
. Find
.
3x  5
dx
1. Let y 
Assume:
u ( x)  x  2 , u( x)  1 and
v( x)  3x  5, v( x)  3
Then, y( x) 
dy (3x  5)  3( x  2)
11


dx
(3x  5)2
(3x  5)2
2. Let y  tan 
sin 
dy
. Find
.
cos
d
dy cos  cos  sin   ( sin  )
1


d
cos2 
cos2 
 sec2 
3. Let y 
sin 

dy  cos  sin 

d
2
4. A certain truck depreciates according to the formula
V
60,000
1  0.3t  0.1t 2
V is measured in $, t is in time years. t  0 is the time
when the vehicle is purchased. Find the rates of its
depreciation (a) at 2 years, (b) at 4 years.
dV  60000  (0.3  0.2t )

dt
(1  0.3t  0.1t 2 )2
dV
0.7  60000

 $10,500 per year
dt
4
dV
66000
(b) at t  4,

 $4571 per year
dt
14.44
(a) at t  2,
Chain rule.
If y is function of u , and u is function of x then
dy dy du

dx du dx
 3x 
Example. Let y  

4 x
3
If we let u 
Note that
and
3x
then y  u 3
4 x
du 3(4  x)  3x  1
12


dx
(4  x) 2
(4  x) 2
dy
 3u 2
du
Therefore, by chain rule,
dy
12
 3u 2 
dx
(4  x) 2

=
2. y  e
27 x 2
(4  x)

2
12
(4  x) 2
324 x 2
(4  x) 4
x
Let u   x   x1/ 2 . Then y  eu
dy
du
1
1
  x 1/ 2  
 eu and
dx
2
2 x
du
dy
e x
u 1
 e

Therefore,
dx
2 x
2 x
3. y  sin 4  . Let u  sin  . Then y  u 4
Then
du
dy
 cos and
 4u 3
d
du
Therefore,
dy
 4 sin 3  cos
d
t t
4. y  a
. Let u  t t . Then u 2  t 3 . Differentiating
both sides with respect to t, we get
du
du 3t 2
3t
3
2
2u
 3t and this yields,



t
dt
dt 2u 2 t 2
Since y  au , we can take log both sides and then
differentiate,
y  au  ln y  u ln a
Differentiating this, we get
Or,
1 dy
 ln a
y du
dy
 y ln a
du
Therefore,
dy 3
 y ln a t
dt 2
More problems in chain-rule based differentiations.
Differentiate the following.
x3
1. y  cos x 2. y  sin(ln x ) 3. y  e
4. y  ( 1  x3 )3 / 4 5. y  sin( 5 x4  3 x  2 )
1
6. y 
sin( 5 x4  3 x  2 )
2