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H = f(g(x 0)∆g)−f(g(x 0)) = f(g ∆g)−f(g) Thus we apply the fundamental lemma of differentiation, h = f0(g)η(∆g)∆g, 1 f0(g)η(∆g) ∆g h Note that f0(g(x)) > 0 for all x ∈ (a,b) and η(∆g) → 0 as h → 0, thus, lim h→0 ∆g/h = lim h→0 1 f0(g)η(∆g) 1 f0(g(x)) Thus g0(x) = 1 f0(g(x)), g 0(f(x)) = 1 f0(x) 3 Suppose g is a real function on R1, with boundedCalculus tells us that if F(x) is theantiderivativeof a function f(x) on the interval a;b, then I = Z b a f(x) dx = F(x)jb a = F(b) F(a) Practically, most integrals cannot be evaluated using this approach For example, Z 1 0 dx 1 x5 has a complicated antiderivative and it easier to adopt a numerical method to approximate this integral Numerical Analysis Trapezoidal and116 = H > B R G B D g Z F b g g h _ h e h ` d b y m g b \ _ j k b l _ l " K \ B \ Z g J b e k d b", L h f 53, K \I 1 1, F _ o Z g b a Z p b y, _ e _ d l j b n Ueltschi Org ƒEƒ‹ƒt ƒ~ƒfƒBƒAƒ€ "¯Œ^ ƒŒƒfƒB[ƒX