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ch02_05

# ch02_05 - 2-45 SECTION 2.5 The Chain Rule 189 2.5 THE CHAIN...

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2-45 SECTION 2.5 . . The Chain Rule 189 2.5 THE CHAIN RULE Suppose that the function P ( t ) = 100 + 8 t models the population of a city after t years. Then the rate of growth of that population after 2 years is given by P ± (2). At present, we need the limit deﬁnition to compute this derivative. However, observe that P ( t )i s the composition of the two functions f ( t ) = t and g ( t ) = 100 + 8 t ,so that P ( t ) = f ( g ( t )). Also, notice that both f ± ( t ) and g ± ( t ) are easily computed using existing derivative rules. We now develop a general rule for the derivative of a composition of two functions. The following simple examples will help us to identify the form of the chain rule. Notice that from the product rule d dx [( x 2 + 1) 2 ] = d [( x 2 + 1)( x 2 + 1)] = 2 x ( x 2 + 1) + ( x 2 + 1)2 x = 2( x 2 + 1)2 x . Of course, we can write this as 4 x ( x 2 + 1), but the unsimpliﬁed form helps us to understand the form of the chain rule. Using this result and the product rule, notice that d [( x 2 + 1) 3 ] = d [( x 2 + 1)( x 2 + 1) 2 ] = 2 x ( x 2 + 1) 2 + ( x 2 + 1)2( x 2 + 1)2 x = 3( x 2 + 1) 2 2 x . We leave it as a straightforward exercise to extend this result to d [( x 2 + 1) 4 ] = 4( x 2 + 1) 3 2 x . You should observe that, in each case, we have brought the exponent down, lowered the power by one and then multiplied by 2 x , the derivative of x 2 + 1 . Notice that we can write ( x 2 + 1) 4 as the composite function f ( g ( x )) = ( x 2 + 1) 4 , where g ( x ) = x 2 + 1 and f ( x ) = x 4 . Finally, observe that the derivative of the composite function is d [ f ( g ( x ))] = d [( x 2 + 1) 4 ] = 4( x 2 + 1) 3 2 x = f ± ( g ( x )) g ± ( x ) . This is an example of the chain rule, which has the following general form. THEOREM 5.1 (Chain Rule) If g is differentiable at x and f is differentiable at g ( x ), then d [ f ( g ( x ))] = f ± ( g ( x )) g ± ( x ) .

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190 CHAPTER 2 . . Differentiation 2-46 PROOF At this point, we can prove only the special case where g ± ( x ) ²= 0. Let F ( x ) = f ( g ( x )). Then, d dx [ f ( g ( x ))] = F ± ( x ) = lim h 0 F ( x + h ) F ( x ) h = lim h 0 f ( g ( x + h )) f ( g ( x )) h Since F ( x ) = f ( g ( x )). = lim h 0 f ( g ( x + h )) f ( g ( x )) h g ( x + h ) g ( x ) g ( x + h ) g ( x ) Multiply numerator and denominator by g ( x + h ) g ( x ). = lim h 0 f ( g ( x + h )) f ( g ( x )) g ( x + h ) g ( x ) lim h 0 g ( x + h ) g ( x ) h Regroup terms. = lim g ( x + h ) g ( x ) f ( g ( x + h )) f ( g ( x )) g ( x + h ) g ( x ) lim h 0 g ( x + h ) g ( x ) h = f ± ( g ( x )) g ± ( x ) , where the next to the last line is valid since as h 0, g ( x + h ) g ( x ), by the continuity of g . (Recall that since g is differentiable, it is also continuous.) You will be asked in the exercises to ﬁll in some of the gaps in this argument. In particular, you should identify why we need g ± ( x ) ²= 0in this proof. It is often helpful to think of the chain rule in Leibniz notation. If y = f ( u ) and u = g ( x ), then y = f ( g ( x )) and the chain rule says that dy = du . (5.1) REMARK 5.1 The chain rule should make sense intuitively as follows. We think of as the (instantaneous) rate of change of y with respect to x , as the (instantaneous) rate of change of y with respect to u and as the (instantaneous) rate of change of u with respect to x .
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ch02_05 - 2-45 SECTION 2.5 The Chain Rule 189 2.5 THE CHAIN...

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