Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

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)
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