This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: (3/19/08) Section 2.7 Derivatives of powers of functions Overview: In this section we discuss the Chain Rule formula for the derivatives of composite functions that are formed by taking powers of other functions. This is a special case of the general Chain Rule which we will cover in Section 4.1. Topics: • The Chain Rule for powers of functions • On the order of operations The Chain Rule for powers of functions How is the rate of change of the area A = w 2 of the square in Figure 1 determined by its width w = w ( t ) and the rate of change of its width? We can view the area as the product A = w · w and use the Product Rule from the last section to obtain dA dt = d dt ( w 2 ) = d dt ( w · w ) = w dw dt + w dw dt = 2 w dw dt . (1) This shows that the rate of change of the area equals twice the width, multiplied by the rate of change of the width. If we replace w = w ( t ) by f = f ( x ) in (1) , we obtain d dx ( f 2 ) = 2 f df dx . This is the special case for n = 2 of the following general result: w = w ( t ) w = w ( t ) FIGURE 1 Theorem 1 (The Chain Rule for powers of functions) Suppose that n is a constant and that y = f ( x ) is a function of x . Then d dx ( f n ) = nf n- 1 df dx (2a) or with primes denoting x-derivatives, ( f n ) prime = nf n- 1 f prime . (2b) This formula holds at any x such that f prime = f prime ( x ) exists and f = f ( x ) is in an open interval where [ f ( x )] n- 1 is defined. Remember (2a) and (2b) as the following statement: the derivative of the n th power of a function equals n , multiplied by the ( n- 1)st power of the function, multiplied by the derivative of the function. 149 p. 150 (3/19/08) Section 2.7, Derivatives of powers of functions Proof: Suppose that x satisfies the conditions of the theorem. For nonzero Δ x , we let Δ f denote the change f ( x + Δ x )- f ( x ) in the value of f from x to x + Δ x , so that f ( x + Δ x ) = f ( x ) + Δ f . Then Δ f → 0 as Δ x → 0 since f is continuous at x . We make the simplifying assumption that Δ f negationslash = 0 for small, nonzero Δ x . Then the derivative of y = f n at x is the limit as Δ x → 0 of [ f ( x + Δ x )] n- [ f ( x )] n Δ x = ( f + Δ f ) n- f n Δ x = bracketleftbigg ( f + Δ f ) n- f n Δ f bracketrightbiggbracketleftbigg Δ f Δ x bracketrightbigg . (3) We have written f here for f ( x ) and we obtained the last expression by multiplying and dividing by Δ f . We let Δ x tend to zero. The difference quotient in the second set of brackets on the right of (3) tends to the derivative of f with respect to x and the difference quotient in the first set of brackets tends to the derivative of f n with respect to f . The difference quotient on the left of (3) therefore tends to the derivative of f n with respect to x , and we obtain d df ( f n ) = d df ( f n ) df dx ....
View Full Document