91 (iii) Product Rule. The function fg is differentiable at x0 and ( fg) ′ (x0)= f ′ (x0)g(x0)+ f (x0)g ′ (x0). f (iv) Quotient Rule. Suppose additionally that g(x0) ̸ = 0. Then the function is differentiable at g x0 and ′ f f ′ (x0)g(x0) − f (x0)g ′ (x0) (x0)= . g (g(x0)) 2 Proof: The proofs of (i) and (ii) are straightforward and we leave them as exercises. We prove (iii). For every x ∈ I \{x0}, we can write ( fg)(x) − ( fg)(x0) f (x)g(x) − f (x0)g(x)+ f (x0)g(x) − f (x0)g(x0) = x − x0 x − x0 ( f (x) − f (x0))g(x) f (x0)(g(x) − g(x0)) = + . x− x0 x − x0 By Theorem 4.1.1, the function g is continuous at x0 and, hence, limx →x0 g(x)= g(x0). Thus, ( fg)(x) − ( fg)(x0) lim = f ′ (x0)g(x0)+ f (x0)g ′ (x0) x→x0 x − x0 and (iii) follows. Next we prove (iv). Since g(x0) ̸ = 0, and g is continuous at x0, it follows from Lemma 3.4.3 or Remark 3.4.1 that there exists an open interval I′ ⊂ I containing x0 such that g(x) ̸ = 0 for all x ∈ I′ . f Let h = . Then h is defned on I′ . Moreover, g h(x) − h(x0) = f (x) f (x0) f (x0) f (x0) − + − g(x) g(x) g(x) g(x 0) x− x0 x − x0 = 1 f (x0) ( f (x) − f (x0)) + (g(x0) − g(x)) g(x) g(x)g(x 0) = x − x0 1 f (x) − f (x0) g(x) − g(x0) g(x0) − f (x0) . g(x)g(x0) x − x0 x − x0 Taking the limit as x → x0, we obtain (iv). The proof is now complete. □ ■ Example 4.1.2 Let f : R → R be 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 → R be given by g(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 → R given by g(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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