Differentials
Here
again, we use the derivative concept to introduce “Differential”. Let us take
another look at (1),
Here df/dx is a notation, used by Leibniz, for
the limit of the difference quotient, which is called the differential coefficient of y with respect to x .Will it be meaningful to treat df/dx as a quotient of df and dx ? In other words, is it possible to assign meaning to df and dx so that derivative is equal to the quotient of df and dx .Well, in some cases yes. For instance, if f ( x) =
mx + c
, m, c are constants, then, y = f (
x) .
Δy = f ( x +
Δx ) − f (x) = mΔx = f ′(x)Δx for all x ∈
ℝ and Δx
and
hence equality in both (2), and (3). In this case changes in x and y (= f ) are taking place along straight
lines, in which case we have,
change
in f / change in x =
Thus in
this case the derivative df/dx is
truly a quotient of df and dx, if we take df = Δf = dy and dx = Δx. This leads us to define the
differential of f as follows:
Let f : (a,
b) → ℝ be a differentiable function, for x ∈(a, b) and Δx the increment given to x , we define the differential of f by
df = f ′
(x)Δx . ... (8)
First we
note that if f (x) = x , then by (8) we get dx = f
′(x)Δ x = 1Δx which means that
the differential dx = Δx
, which is the change in x -axis.
So the differential given by (8) is same as df
= f ′(x)dx .
Linear Approximation and Differential
Next we explore
the differential for an arbitrary differentiable function y = f (x).
Then Δf = f ( x + dx ) − f (x) gives the change in output along the graph of y = f (x) and f ′(x) gives the slope of the tangent line at ( x , f (x)) . Let dy or df denote the increment in f along the tangent line. Then by the above observation, we have dy = f ′(x)dx .
From the
figure it is clear that Δf ≈ dy = df
= f ′(x)dx and hence f ′ (x) can be viewed approximately as the
quotient of Δf and Δx. So we may interpret df/dx as the quotient of df and dx .
We know
that derivative of a function is again a function. On the other hand, differential
df of a function f is not only a function of the independent variable but also
depends on the change in the input namely dx
=
Δx . So df is a function of two changing quantities namely x and dx. Observe that Δf≈ df , which can be observed from the
Fig. 8.4.
In the
table below we give some functions, their derivatives and their differentials
side by side for comparative purpose.
Next we
look at the properties of differentials. These results easily follow from the
definition of differential and the rules for differentiation. We give a proof
for (5) below and the other proofs are left as exercises.
Here we
consider real-valued functions of real variable.
(1) If f is a constant function, then df = 0 .
(2) If f (x) = x identity function, then df
= 1dx .
(3) If f is differentiable and c ∈ R, then d(cf) = cf ′(x)dx .
(4) If f ,
g are differentiable, then d ( f +
g) = df + dg = f ′ (x)dx + g′(x)dx .
(5) If f ,
g are differentiable, then d ( fg ) = fdg + gdf = ( f (x) g ′ (
x) + f ′ (x) g ( x))dx.
(6) If f ,
g are differentiable, then , where g(x)≠0.
(7) If f ,
g are differentiable and h = f o g is
defined, then dh = f ′( g( x )) g′( x ) dx .
(8) If
h(x) = ef (x) , then dh = ef
(x) f ′( x ) dx .
(9) If f (x) > 0 for all x and g (x) = log( f (x)) , then dg = f ′(x)/f(x) dx .
Let f ,
g : (a , b) → ℝ
be differentiable functions. Show that d(fg
) = fdg + gdf .
Let f ,
g : (a , b) → ℝ
be differentiable functions and h(x) = f
(x) g (x) . Then h , being a product of differentiable functions, is
differentiable on (a, b) . So by
definition dh = h′ ( x)dx. Now by using product rule we have h′ (x)
= f (x) g′ ( x) + f ′(x) g(x) .
Thus dh = h′
(x)dx = ( f (x) g′ ( x) + f ′ (x) g( x))dx = f
(x) g′ ( x)dx + f ′ (x) g( x)dx
= f (x)dg + g ( x)df = fdg + gdf
Example 8.6
Let g ( x)
=
x2 +
sin x . Calculate the differential dg .
Solution
Note
that g is differentiable and g ′(x)
=
2x + cos x .
Thus dg = (2x + cos x)dx .
Example 8.7
If the
radius of a sphere, with radius 10 cm, has to decrease by 0.1 cm, approximately
how much will its volume decrease?
Solution
We know
that volume of a sphere is given by V
=
4/3 πr3 , where r > 0 is the radius. So the differential
dV = 4πr2dr and hence
ΔV ≈ dV
= 4π(10)2 (9.9-10) cm3
= 4102 ( 0.1) cm3
= − 40π cm3.
Note
that we have used dr =
(9.9 −10)
cm, because radius decreases from 10 to 9.9. Again the negative sign in the
answer indicates that the volume of the sphere decreases about 40π cm3.
Related Topics
Privacy Policy, Terms and Conditions, DMCA Policy and Compliant
Copyright © 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.