Here again, we use the derivative concept to introduce “Differential”.

**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 *+*
*Δ*x *) − *f (x)* = mΔ*x *=* f *′(*x*)Δ*x *for all* x *∈
ℝ and Δ*x*

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* = Δ*f* = *dy* and *dx *= Δ*x*. 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 Δ*x* the increment given to *x* , we define the differential of *f* by

*df* = *f *′*
*(*x*)Δ*x *. ... (8)

First we
note that if *f (x)* =* x *, then by (8) we get *dx* = *f
*′*(x)*Δ* x *= 1Δ*x *which means that
the differential *dx* =* *Δ*x
*, 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* *Δ*f
*=* 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 Δ*f* ≈ *dy* = *df*
= *f *′*(x)dx* and hence *f *′* (x)* can be viewed approximately as the
quotient of Δ*f* and Δ*x*. 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*
=
Δ*x* . So *df* is a function of two changing quantities namely *x* and *dx*. Observe that Δ*f*≈ *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*
= 1*dx* .

(3) If* f *is differentiable and c ∈ **R**, then *d*(*cf*) = c*f *′*(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*) = e* ^{f (x)}* , then dh = e

(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*)
=
*x*^{2} +
sin *x* . Calculate the differential *dg* .

**Solution**

Note
that *g* is differentiable and *g* ′(*x*)
=
2*x* + cos *x* .

Thus *dg* = (2*x* + 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 *π**r*^{3} , where *r* > 0 is the radius. So the differential
*dV* = 4*π**r*^{2}*dr* and hence

Δ*V *≈* dV
*=* *4*π*(10)^{2}* *(9.9-10) cm^{3}

= 410^{2} ( 0.1) cm^{3}

= − 40π cm^{3}.

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 cm^{3}.

Tags : Mathematics , 12th Maths : UNIT 8 : Differentials and Partial Derivatives

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