121616949-math.67

# 121616949-math.67 - f x/g x if we already know f ± x and g...

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3.4 The Quotient Rule 53 This is similar to the use of P to denote a sum. For example, 5 Y k =1 f k = f 1 f 2 f 3 f 4 f 5 and n Y k =1 k = 1 · 2 · . . . · n = n ! . We sometimes use somewhat more complicated conditions; for example n Y k =1 ,k = j f k denotes the product of f 1 through f n except for f j . For example, 5 Y k =1 ,k =4 x k = x · x 2 · x 3 · x 5 = x 11 . 8. The generalized product rule says that if f 1 , f 2 , . . . , f n are differentiable functions at x then d dx n Y k =1 f k ( x ) = n X j =1 0 @ f j ( x ) n Y k =1 ,k = j f k ( x ) 1 A . Verify that this is the same as your answer to the previous problem when n = 4, and write out what this says when n = 5. What is the derivative of ( x 2 + 1) / ( x 3 - 3 x )? More generally, we’d like to have a formula to compute the derivative of
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Unformatted text preview: f ( x ) /g ( x ) if we already know f ± ( x ) and g ± ( x ). Instead of attacking this problem head-on, let’s notice that we’ve already done part of the problem: f ( x ) /g ( x ) = f ( x ) · (1 /g ( x )), that is, this is “really” a product, and we can compute the derivative if we know f ± ( x ) and (1 /g ( x )) ± . So really the only new bit of information we need is (1 /g ( x )) ± in terms of g ± ( x ). As with the product rule, let’s set this up and see how...
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