Mathematical Methods for Physics and Engineering : A Comprehensive Guide

2.1 DIFFERENTIATION

separation is not unique. (In the given example, possible alternative break-ups

would beu(x)=x^2 ,v(x)=xsinx,orevenu(x)=x^4 tanx,v(x)=x−^1 cosx.)

The purpose of the separation is to split the function into two (or more) parts,

of which we know the derivatives (or at least we can evaluate these derivatives

more easily than that of the whole). We would gain little, however, if we did

not know the relationship between the derivative offand those ofuandv.

Fortunately, they are very simply related, as we shall now show.

Sincef(x) is written as the productu(x)v(x), it follows that

f(x+∆x)−f(x)=u(x+∆x)v(x+∆x)−u(x)v(x)

=u(x+∆x)[v(x+∆x)−v(x)] + [u(x+∆x)−u(x)]v(x).

From the definition of a derivative (2.1),

df

dx

= lim

∆x→ 0

f(x+∆x)−f(x)

∆x

= lim

∆x→ 0

{

u(x+∆x)

[

v(x+∆x)−v(x)

∆x

]

+

[

u(x+∆x)−u(x)

∆x

]

v(x)

}

.

In the limit ∆x→0, the factors in square brackets becomedv/dxanddu/dx

(by the definitions of these quantities) andu(x+∆x) simply becomesu(x).

Consequently we obtain

df

dx

=

d

dx

[u(x)v(x)] =u(x)

dv(x)

dx

+

du(x)

dx

v(x). (2.6)

In primed notation and without writing the argumentxexplicitly, (2.6) is stated

concisely as

f′=(uv)′=uv′+u′v. (2.7)

This is a general result obtained without making any assumptions about the

specific formsf,uandv, other than thatf(x)=u(x)v(x). In words, the result

reads as follows.The derivative of the product of two functions is equal to the

first function times the derivative of the second plus the second function times the

derivative of the first.

Find the derivative with respect toxoff(x)=x^3 sinx.

Using the product rule, (2.6),

d

dx

(x^3 sinx)=x^3

d

dx

(sinx)+

d

dx

(x^3 )sinx

=x^3 cosx+3x^2 sinx.

The product rule may readily be extended to the product of three or more

functions. Considering the function

f(x)=u(x)v(x)w(x) (2.8)