92 4.1 Definition 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 →Randg: J →R, where I andJ are open intervals inR with f (I) ⊂J. Let x0 ∈I and suppose f is differentiable at x0 and gis 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 lim x→x0 g(f (x))−g(f (x0)) x−x0 exists and equals g′(f (x0))f ′(x0). If we could write (g◦ f )(x)−(g◦ f )(x0) x−x0 = g(f (x))−g(f (x0)) x−x0 = g(f (x))−g(f (x0)) f (x)−f (x0) · f (x)−f (x0) x−x0 and take the limit as x →x0, we would get the desired result. Unfortunately, even though x̸ =x0, the expression f (x)−f (x0) could be zero for values of x arbitrarily close tox0. When that is the case, we cannot divide by that difference. To fix this situation, we will introduce an auxiliary functionh: J →Rgiven by h(y) = g(y)−g(f (x0)) y−f (x0) , if y̸ =f (x0); g′(f (x0), if y =f (x0). Note that if y̸ = 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 x̸ =x0, we get the equality g(f (x))−g(f (x0)) x−x0 =h(f (x))· f (x)−f (x0) x−x0 Moreover, since gis differentiable at f (x0), we have that his continuous at f (x0) because lim y→f (x0) h(y) = lim y→f (x0) g(y)−g(f (x0)) y−f (x0) =g′(f (x0)) =h(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 lim x→x0 g(f (x))−g(f (x0)) x−x0 = lim x→x0 h(f (x))· lim x→x0 f (x)−f (x0) x−x0 =h(f (x0))f ′(x0) =g′(f (x0))f ′(x0) and the theorem holds. □
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