Mathematical Methods for Physics and Engineering - Matematica.NET

2.1 DIFFERENTIATIONseparation is not unique. (In the given example, possible alternative break-upswould be u(x) =x 2 , v(x) =x sin x, orevenu(x) =x 4 tan x, v(x) =x −1 cos x.)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 derivativesmore easily than that of the whole). We would gain little, however, if we didnot know the relationship between the derivative of f and those of u and v.Fortunately, they are very simply related, as we shall now show.Since f(x) is written as the product u(x)v(x), it follows thatf(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),dfdx = lim f(x +∆x) − f(x)∆x→0{∆x= lim u(x +∆x)∆x→0 ∆x[ v(x +∆x) − v(x)]+[ u(x +∆x) − u(x)∆x] }v(x) .In the limit ∆x → 0, the factors in square brackets become dv/dx and du/dx(by the definitions of these quantities) and u(x +∆x) simply becomes u(x).Consequently we obtaindfdx = d [u(x)v(x)] = u(x)dv(x)dx dx+ du(x) v(x). (2.6)dxIn primed notation and without writing the argument x explicitly, (2.6) is statedconcisely asf ′ =(uv) ′ = uv ′ + u ′ v. (2.7)This is a general result obtained without making any assumptions about thespecific forms f, u and v, other than that f(x) =u(x)v(x). In words, the resultreads as follows. The derivative of the product of two functions is equal to thefirst function times the derivative of the second plus the second function times thederivative of the first.◮Find the derivative with respect to x of f(x) =x 3 sin x.Using the product rule, (2.6),ddx (x3 sin x) =x 3 d d(sin x)+dx dx (x3 )sinx= x 3 cos x +3x 2 sin x. ◭The product rule may readily be extended to the product of three or morefunctions. Considering the functionf(x) =u(x)v(x)w(x) (2.8)45