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Differential
of a function
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Use
of differential to approximate the value of a function
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If
the change of the independent variable (increment)
Dx
is
small then, the differential dy
of a function
y = f (x)
and
the increment
Dy
are approximately equal, |
|
Dy
»
dy
that is, f (x + Dx)
-
f (x)
= f '(x)Dx
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|
or
f (x + Dx)
»
f (x)
+ f '(x)Dx. |
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| Example:
Using
the differential calculate approximate value of e
0.15.
|
| Solution:
Using
the formula
|
| f (x
+ Dx)
»
f (x)
+ f '(x)Dx
and
taking,
f (x)
= ex
where,
x
= 0
and Dx
= 0.15 |
|
then, ex
+ Dx
»
ex
+ (ex)'
· Dx
that
is,
e 0.15
»
e
0
+ e
0
· 0.15
= 1 + 0.15
= 1.15. |
|
| Example:
Approximately,
how much will increase the volume of the sphere if its
radius r
= 20
cm increases
by 2 mm.
|
| Solution:
Using
approximation Dy
»
dy
or Dy
= f(x + Dx)
-
f(x)
»
f '(x) Dx
.
|
| Therefore,
DV
»
V '(r) Dr
and
since V
= (4/3)pr3
then, V '(r)
= (4/3)p
· 3r2
= 4pr2. |
| By substituting
r
= 20
and Dr
= 0.2
into |
| DV
»
V '(r)Dr
= 4pr2
· Dx
we get
DV
»
4p
· 202
· 0.2
= 1005,3
cm3. |
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Rules
for differentials
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| Rules
for differentials are the same to those for derivatives, such that |
| 1)
dc = 0,
c
is a constant |
| 2)
dx = Dx,
x
is the independent variable |
| 3)
d(cu) = c
du |
| 4)
d(u ±
v) = du ±
dv |
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5)
d(u v) = u
dv ±
v du |
|
6) |
 |
|
7) |
d
f(u) = f ' (u) du. |
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Differentials
of some basic functions
|
| 1)
|
d
un = n
un -
1du |
| 2)
|
d
eu = eu
du |
| 3)
|
d
an = an ln a
du |
| 4)
|
 |
| 5)
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d
sin u = cos u
du |
| 6)
|
d
cos u = -
sin u
du |
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| Example: To
show the use of the formula 2)
d
eu = eu du
above, let substitute;
a) u =
x,
b) u = cos
x
and
c) u =
x3. |
| Solution:
a)
By substituting u =
x,
obtained is d
ex = ex dx |
|
b) By substituting u =
cos x,
we get d
ecos x = ecos x
d(cos x) =
-
ecos x sin x
dx |
|
c)
By substituting u =
x3,
we get |
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