2.3. The Derivative 65
- Show that the first two terms in the expansion of a polynomial
p(t) = a&” + a,-#-l + * * * + a1t + a0in terms of ascending powers of (t - c) areP(C) + p’(c)(t - c)
where p’(t) is the polynomial p’(t) = na,t”-’ + (n - l)an-$‘-2 +
(n - 2)a,-2iT3 +... + 2a2t + al.- The derivative. For any polynomial p(t), the polynomial p’(t) defined
in Exercise 3 is called the derivative of p(t). The process of obtaining
the derivative of a polynomial is called diflerentiation.
(a) Verify that the derivative of the polynomial 3t5-7t4+6t2-5t-3
is 15t4 - 28t3 + 12t - 5.
(b) Find the derivative of the polynomial 4t13 - 3ta - 5t7+4t3 + 76t.- Properties of the derivative. Establish the following properties of the
derivative, where p and q are polynomials and k is a constant:
(a) (p + q)‘(t) = p’(t) + q’(t). Extend this to an arbitrary sum of
polynomials.
(b) W’(t) = W(t).
cc> $J$.lys= p’(t)q(t) + p(t)q’(t) in the special case p(t) = tp and
(4 W’(t) 1 p’(Mt) +p(th’(t) f or arbitrary polynomials p and q.
Extend this to a product of more than two polynomials.
(e) the derivative of p(t)’ is rp(t)‘-‘p’(t) for an arbitrary positive
integer r (use (d)).
(f) (P 0 q)‘(t) = p’(q(t))q’(t).- Verify the properties of the derivative given in Exercise 5 for the
special case
p(t) = 3t2 - 4t + 2 and q(t) = 4t3 - 2t2 + 6t + 1.- Since differentiation of a polynomial leads to another polynomial, we
can apply the operation of taking the derivative repeatedly. Thus, if
p’(t) is the derivative of p(t), th e second derivative of p(t) is p”(t) =
(p’)‘(t). In general, we can define
p(O)(t) = p(t)
p(l)(t) = p’(t)
p’“‘(t) = p”(t)
pck)(t) = (p(k-‘))‘(t) for k > - 3.