93 ■ Example 4.1.3 Consider the function h: R → R given by h(x)=(3x 4 + x + 7)15. Since h(x) is a polynomial we could in principle compute h′ (x) by expanding the power and using Example 4.1.2. However, Theorem 4.1.3 provides a shorter way. Defne f ,g : R → R by f (x)= 3x4 + x + 7 and g(x)= x15. Then h = g ◦ f . Given x 0 ∈ R, it follows from Theorem 4.1.3 that 4 3 (g ◦ f ) ′ (x 0)= g ′ ( f (x0)) f ′ (x0)= 15(3x0 + x0 + 7) 14(12x 0 + 1). ■ Example 4.1.4 By iterating the Chain Rule, we can extended the result to the composition of more than two functions in a straightforward way. For example, given functions f : I1 → R, g: I2 → R, and h: I3 → R such that f (I1) ⊂ I2, g(I2) ⊂ I3, f is differentiable at x0, g is differentiable at f (x0), and h is differentiable at g( f (x0)), we obtain that h ◦ g ◦ f is differentiable at x0 and (h ◦ g ◦ f ) ′ (x0)= h′ (g( f (x0)))g ′ ( f (x0)) f ′ (x0). Defnition 4.1.2 Let I be an open interval in R and let f : I → R be a differentiable function. If the function f ′ : I → R is also differentiable, we say that f is twice differentiable (on I). The second ′′ ′′ derivative of f is denoted by f or f (2). Thus, f =( f ′ ) ′ . Similarly, we say that f is three times differentiable if f (2) is differentiable, and ( f (2)) ′ is called the third derivative of f and is denoted ′′′ by f or f (3). We can defne in this way n times differentiability and the nth derivative of f for any positive integer n. As a convention, f (0) = f . Defnition 4.1.3 Let I be an open interval in R and let f : I → R. The function f is said to be ′ continuously differentiable on I if f is differentiable on I and f is continuous on I. We denote by C1(I) the set of all continuously differentiable functions on I. If f is n times differentiable on I and the nth derivative is continuous, then f is called n times continuously differentiable on I. We denote by Cn(I) the set of all n times continuously differentiable functions on I. Exercises 4.1.1 Prove parts (i) and (ii) of Theorem 4.1.2. 4.1.2 Compute the following derivatives directly from the defnition. That is, do not use Theorem 4.1.2, but rather compute the appropriate limit directly (see Example 4.1.1). (a) f (x)= mx+ b where m,b ∈ R. 1 (b) f (x)= (here assume x ≠ 0). x √ (c) f (x)= x (here assume x > 0). 4.1.3 Let f : R → R be given by ( x2 , if x > 0; f (x)= 0, if x ≤ 0. (a) Prove that f is differentiable at 0. Find f ′ (x) for all x ∈ R. ′ ′ (b) Is f continuous? Is f differentiable? 4.1.4 Let ( xα , if x > 0; f (x)= 0, if x ≤ 0.
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