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Math Help - differentiable products

  1. #1
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    differentiable products

    Suppose that f: I -> R and g: I -> R are differentiable at c element of I and that g(c) is not equal to 0

    A) show that (1/g)'= -g'(c)/[g(c)]^2
    B) Use part (A) and the product rule to derive the quotient rule
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    Quote Originally Posted by slowcurv99 View Post
    Suppose that f: I -> R and g: I -> R are differentiable at c element of I and that g(c) is not equal to 0

    A) show that (1/g)'= -g'(c)/[g(c)]^2
    Let g(x) be differenciable at c. And g(c)=0.

    Then we can define h(x) = 1/g(x) for x!=c on some open interval containing c. (See proof below).

    Proof: Since g(c) is not zero there is an interval containing c so that g(x)!=0 except at c. The reason is because g(x) is differenciable on I so it is continous on I. Thus, |g(x)-g(c)| < e. Thus, g(c) - e< g(x) < g(c) + e. If g(c) >0 then choose e small enough so that g(c) - e>0. And if g(c) < 0 choose e small enough so that g(c) + e<0. On the interval |x-c|< d so c-d<x<c+d. So choose (c-d,c+d) as your interval.
    Q.E.D.

    This tells us that 1/g(x) is defined on an open interval containing c except possible at c.

    We can therefore define h(x) = (1/x) o (g(x)) and use the chain rule.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by slowcurv99 View Post
    B) Use part (A) and the product rule to derive the quotient rule
    (f(x)/g(x))' = (f(x) * 1/g(x))'

    ...............= f'(x)(1/g(x)) + f(x)(1/g(x))' by the product rule


    but from (A), (1/g(x))' = -g'(x)/[g(x)]^2


    so (f(x)/g(x))' = f'(x)(1/g(x)) + f(x){-g'(x)/[g(x)]^2}
    ....................= f'(x)(1/g(x)) f(x)g'(x)/[g(x)]^2
    ....................= f'(x)/g(x) - f(x)g'(x)/[g(x)]^2 ..............Now combine the fractions
    ....................= [f'(x)g(x) f(x)g'(x)]/g(x)^2 ...............The Quotient Rule
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