Unformatted text preview: f ( x ) /g ( x ) if we already know f ± ( x ) and g ± ( x ). Instead of attacking this problem headon, 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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 Spring '07
 JonathanRogawski
 Math, Calculus, Derivative, Product Rule, Quotient Rule

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