92 4.1 Defnition and Basic Properties of the Derivative We now discuss the differentiability of the composition of functions. Theorem 4.1.3 — Chain rule. Let f : I → R and g: J → R, where I and J are open intervals in R with f (I) ⊂ J. Let x0 ∈ I and suppose f is differentiable at x0 and g is differentiable at f (x0). Then the function g ◦ f is differentiable at x0 and (g ◦ f ) ′ (x0)= g ′ ( f (x0)) f ′ (x0). Proof: Our goal is to prove that g( f (x)) − g( f (x0)) lim x→x0 x − x0 exists and equals g ′ ( f (x0)) f ′ (x0). If we could write (g ◦ f )(x) − (g ◦ f )(x0) g( f (x)) − g( f (x0)) g( f (x)) − g( f (x0)) f (x) − f (x0) = = · x − x0 x − x0 f (x) − f (x0) x − x0 and take the limit as x → x0, we would get the desired result. Unfortunately, even though x ̸ = x0, the expression f (x) − f (x0) could be zero for values of x arbitrarily close to x0. When that is the case, we cannot divide by that difference. To fx this situation, we will introduce an auxiliary function h: J → R given by g(y) − g( f (x0)) , if y ̸ = f (x0); h(y)= y− f (x0) g ′ ( f (x0), if y = f (x0). Note that if y ≠ f (x0), then g(y) − g( f (x0)) = h(y)(y − f (x0)) and if y = f (x0), then g(y) − g( f (x0)) = 0 = h(y)(y − f (x0)). Hence, g(y) − g( f (x0)) = h(y)(y − f (x0)) for all y ∈ J. Since f (I) ⊂ J we obtain g( f (x)) − g( f (x0)) = h( f (x))( f (x) − f (x0)), for all x ∈ I. Now, for x ̸ = x0, we get the equality g( f (x)) − g( f (x0)) f (x) − f (x0) = h( f (x)) · x − x0 x − x0 Moreover, since g is differentiable at f (x0), we have that h is continuous at f (x0) because g(y) − g( f (x0)) lim h(y)= lim = g ′ ( f (x0)) = h( f (x0)). y→ f (x0) y→ f (x0) y − f (x0) Since f is differentiable at x0, f is also continuous at x0. Therefore, the composition h ◦ f is continuous at x0. Putting this information together, we have g( f (x)) − g( f (x0)) f (x) − f (x0) lim = lim h( f (x)) · lim = h( f (x0)) f ′ (x0)= g ′ ( f (x0)) f ′ (x0) x→x0 x − x0 x→x0 x→x0 x − x0 and the theorem holds. □
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