Section2_7 - Section 2.7 Derivatives of powers of functions...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the 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

{[ snackBarMessage ]}

Page1 / 8

Section2_7 - Section 2.7 Derivatives of powers of functions...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online