121616949-math.72

# 121616949-math.72 - rule The function is(625-x 2-1 2 the...

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58 Chapter 3 Rules For Finding Derivatives This looks like trivial arithmetic, but it is not: dg/dx is not a fraction, that is, not literal division, but a single symbol that means g ( x ). Nevertheless, it turns out that what looks like trivial arithmetic, and is therefore easy to remember, is really true. It will take a bit of practice to make the use of the chain rule come naturally—it is more complicated than the earlier differentiation rules we have seen. EXAMPLE 3.9 Compute the derivative of 625 - x 2 . We already know that the answer is - x/ 625 - x 2 , computed directly from the limit. In the context of the chain rule, we have f ( x ) = x , g ( x ) = 625 - x 2 . We know that f ( x ) = (1 / 2) x - 1 / 2 , so f ( g ( x )) = (1 / 2)(625 - x 2 ) - 1 / 2 . Note that this is a two step computation: first compute f ( x ), then replace x by g ( x ). Since g ( x ) = - 2 x we have f ( g ( x )) g ( x ) = 1 2 625 - x 2 ( - 2 x ) = - x 625 - x 2 . EXAMPLE 3.10 Compute the derivative of 1 / 625 - x 2 . This is a quotient with a
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Unformatted text preview: rule. The function is (625-x 2 )-1 / 2 , the composition of f ( x ) = x-1 / 2 and g ( x ) = 625-x 2 . We compute f ± ( x ) = (-1 / 2) x-3 / 2 using the power rule, and then f ± ( g ( x )) g ± ( x ) =-1 2(625-x 2 ) 3 / 2 (-2 x ) = x (625-x 2 ) 3 / 2 . In practice, of course, you will need to use more than one of the rules we have developed to compute the derivative of a complicated function. EXAMPLE 3.11 Compute the derivative of f ( x ) = x 2-1 x √ x 2 + 1 . The “last” operation here is division, so to get started we need to use the quotient rule ﬁrst. This gives f ± ( x ) = ( x 2-1) ± x √ x 2 + 1-( x 2-1)( x √ x 2 + 1) ± x 2 ( x 2 + 1) = 2 x 2 √ x 2 + 1-( x 2-1)( x √ x 2 + 1) ± x 2 ( x 2 + 1) ....
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