91 (iii) Product Rule. The function fgis differentiable at x0 and (fg)′(x0)=f ′(x0)g(x0)+f (x0)g′(x0). (iv) Quotient Rule. Suppose additionally that g(x0)̸=0. Then the function f g is differentiable at x0 and f g ′ (x0)= f ′(x0)g(x0)−f (x0)g′(x0) (g(x0)) 2 . Proof: The proofs of (i) and (ii) are straightforward and we leave them as exercises. We prove (iii). For everyx ∈I \{x0}, we can write (fg)(x)−(fg)(x0) x−x0 = f (x)g(x)−f (x0)g(x)+f (x0)g(x)−f (x0)g(x0) x−x0 = (f (x)−f (x0))g(x) x−x0 + f (x0)(g(x)−g(x0)) x−x0 . By Theorem 4.1.1, the functiong is continuous at x0 and, hence, limx →x0 g(x)=g(x0). Thus, lim x→x0 (fg)(x)−(fg)(x0) x−x0 =f ′(x0)g(x0)+f (x0)g′(x0) and (iii) follows. Next we prove (iv). Since g(x0)̸=0, andgis continuous at x0, it follows from Lemma 3.4.3 or Remark 3.4.1 that there exists an open interval I′ ⊂I containingx0 such that g(x)̸=0 for all x ∈I′. Let h= f g . Then his defined on I′. Moreover, h(x)−h(x0) x−x0 = f (x) g(x) − f (x0) g(x) + f (x0) g(x) − f (x0) g(x0) x−x0 = 1 g(x)(f (x)−f (x0))+ f (x0) g(x)g(x0) (g(x0)−g(x)) x−x0 = 1 g(x)g(x0) g(x0) f (x)−f (x0) x−x0 − f (x0) g(x)−g(x0) x−x0 . Taking the limit as x →x0, we obtain (iv). The proof is now complete. □ ■ Example 4.1.2 Let f : R→Rbe given by f (x)=x 2 and let x 0 ∈R. Using Example 4.1.1(a) and the product rule (Theorem 4.1.2(iii)) we can provide an alternative derivation of a formula for f ′(x0). Indeed, let g: R→Rbe givenbyg(x)=x. Then f =g·g so f ′(x0)=(gg)′(x0)=g′(x0)g(x0)+g(x0)g′(x0)=2g′(x0)g(x0)=2x0. Therefore, f ′(x)=2x, for x ∈R. Proceeding by induction, we can obtain the derivative of g: R→Rgivenbyg(x)=xn for n∈N as g′(x)=nxn−1. Furthermore, using this and Theorem 4.1.2(i)(ii) we obtain the familiar formula for the derivative of a polynomial p(x)=anx n +···+a1x+a0 as p′(x)=nanx n−1 +···+2a2x+a1.
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