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Math Help - Backwards Derivative Rules

  1. #1
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    Backwards Derivative Rules

    I came up with two problems which I find interesting.

    1)If f(x)+g(x) is differenciable at c, and f(x) is differenciable at c then is g(x) differenciable at c?

    2)If f(x)g(x) is differenciable at c, and f(x) is differenciable at c then is g(x) differenciable at c?
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I came up with two problems which I find interesting.

    1)If f(x)+g(x) is differenciable at c, and f(x) is differenciable at c then is g(x) differenciable at c?

    2)If f(x)g(x) is differenciable at c, and f(x) is differenciable at c then is g(x) differenciable at c?
    The first is true, for obvious reasons.
    (f + g)' = f' + g'
    If f(x)+g(x) is differentiable at c then f and g must be defined and continuous at c, and therefore, each is independently differentiable at c.

    The second may not be as easy to prove:
    If f(x)*g(x) is differentiable at c and f(x) is differentiable at c, then does g(x) necessarily have to be?
    (f*g)' = f'g + fg'
    This shows that f(x) times g'(x) exists at c. I think it can be proven that this is not true, but I'll just provide a counter example:

    Let f(x) = x^3 and g(x) = 1/x, and c = 0

    (f(x)*g(x)) = x^3*1/x = x^2
    (f(x)*g(x))' = 3x^2*1/x + x^3*-1/x^2 = 3x - x = 2x
    (f(x)*g(x))'(0) = 0
    g'(x) = -1/x^2
    g'(0) = undefined!
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  3. #3
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    Quote Originally Posted by ecMathGeek View Post
    The first is true, for obvious reasons.
    (f + g)' = f' + g'
    If f(x)+g(x) is differentiable at c then f and g must be defined and continuous at c, and therefore, each is independently differentiable at c.
    You are right! It is true, but how do you prove it? It really is easy.

    The second may not be as easy to prove:
    If f(x)*g(x) is differentiable at c and f(x) is differentiable at c, then does g(x) necessarily have to be?
    (f*g)' = f'g + fg'
    This shows that f(x) times g'(x) exists at c. I think it can be proven that this is not true, but I'll just provide a counter example:

    Let f(x) = x^3 and g(x) = 1/x, and c = 0

    (f(x)*g(x)) = x^3*1/x = x^2
    (f(x)*g(x))' = 3x^2*1/x + x^3*-1/x^2 = 3x - x = 2x
    (f(x)*g(x))'(0) = 0
    g'(x) = -1/x^2
    g'(0) = undefined!
    Oh no! Very bad, very bad conter example.

    Given f and g to be two real functions we domains dom(f) and dom(g) respectively. We define the function:
    f * g to be f(x)*g(x) with domain, dom(f) (intersect) dom(g) (hoping the intersection is non-empty).

    So given, f(x)=x^3 on R and given g(x)=1/x on R-{0} then their product is (f*g)(x)=x^2 for R-{0}. Thus, your reasoning does not work. In high school precalculus what you did was okay, but not if you want to be totally mathematically correct.
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  4. #4
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Oh no! Very bad, very bad conter example.

    Given f and g to be two real functions we domains dom(f) and dom(g) respectively. We define the function:
    f * g to be f(x)*g(x) with domain, dom(f) (intersect) dom(g) (hoping the intersection is non-empty).

    So given, f(x)=x^3 on R and given g(x)=1/x on R-{0} then their product is (f*g)(x)=x^2 for R-{0}. Thus, your reasoning does not work. In high school precalculus what you did was okay, but not if you want to be totally mathematically correct.
    If f(x) is continuous and differentiable over some domain D(f) containing x = c (and f(x) != 0), and if f(x)*g(x) is continuous and differentiable over some domain D(f*g) containing x = c, where D(f*g) is the intersection of D(f) and D(g), then D(g) must also be continuous at x = c.

    For f(x) to be continuous at c, f(c) = C is a constant, real number solution.
    For f(x) to be differentiable and g(x) to be continuous at c, f'(c)g(c) = R is some constant, real number solution.
    For (f(x)*g(x)) to be differentiable at c, f'(c)g(c) + f(c)g'(c) = Q is a constant, real number solution.

    Therefore
    (f(x)*g(x))'(c) = f'(c)g(c) + f(c)g'(c) = R + Cg'(c) = Q --> g'(c) = (Q - R)/C is some constant, real number solution if C = f(c) != 0.

    However, if C = f(x) = 0, then R = Q and R - Q = 0, and so
    g'(c) = lim{x->c} (Q - R)/C
    = lim{x->c} (f'(x)g(x) + f(x)g'(x) - f'(x)g(x))/f(x)

    I'm not sure where to go from here if C = 0 (if I'm even going in the right direction). I've tried a few different routes, but I keep getting stuck.
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  5. #5
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    1)If f(x)+g(x) is differenciable at c, and f(x) is differenciable at c then is g(x) differenciable at c?
    If f(x) is continuous and differentiable over some domain D(f) containing x = c and f(x)+g(x) is continuous and differentiable over some domain D(f+g) containing x = c, where D(f+g) is the intersection of D(f) and D(g), then g is continuous at x = c.

    For f(x) to be differentiable at c, f'(c) = C is a constant, real number solution.
    For f(x)+g(x) to be differentiable at c, f'(c)+g'(c) = Q is a constant, real number solution.

    Therefore,
    [f(x) + g(x)]' = f'(x) + g'(x) = C + g'(x) = Q --> g'(x) = Q - C is some constant, real number solution.
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  6. #6
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    Let me post the solutions to show you how easy they really are:

    1)If f(x)+g(x) is differenciable at c and f(x) is differenciable at c then [f(x)+g(x) - f(x)] is differenciable at c. Thus, g(x) is differenciable at c.

    2)If f(x)g(x) is differenciable at c and f(x) is differenciable at c and f(c)!=0 then f(x)g(x)/f(x) is differenciable at c, hence g(x) is differenciable at c.
    But if f(x) is differenciable at c and f(c)=0 then it is not true.
    For example f(x)=x and g(x) = |x| and c=0.
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  7. #7
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Let me post the solutions to show you how easy they really are:

    1)If f(x)+g(x) is differenciable at c and f(x) is differenciable at c then [f(x)+g(x) - f(x)] is differenciable at c. Thus, g(x) is differenciable at c.

    2)If f(x)g(x) is differenciable at c and f(x) is differenciable at c and f(c)!=0 then f(x)g(x)/f(x) is differenciable at c, hence g(x) is differenciable at c.
    But if f(x) is differenciable at c and f(c)=0 then it is not true.
    For example f(x)=x and g(x) = |x| and c=0.
    I wrote more but I pretty much had the same conclusions.
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  8. #8
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    Here is an easy counterexample for #2.
    Consider the case of f(x)=x and g(x)=1/(1+|x|).
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  9. #9
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Plato View Post
    Here is an easy counterexample for #2.
    Consider the case of f(x)=x and g(x)=1/(1+|x|).
    Are you sure that works?
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  10. #10
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    Did you graph x/(1+|x|)?
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  11. #11
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Plato View Post
    Did you graph x/(1+|x|)? Consider x=0.
    Think about the derivatives on each side of x=0.
    The question is, is f(x)*g(x) differentiable at 0. I know g(x) isn't, but I don't think f(x)*g(x) is either.
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  12. #12
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    Say that p(x)=x/(1+|x|) then p(0+h)= h/(1+|h|), p(0)=0, &
    [p(0+h)-p(0)]/h =1/(1+|h|).

    P.S. The function p is a very interesting example. It has a derivative at 0 but at the same time the second derivative does not exist. Nonetheless, the point (0,0) is a point of inflection for the graph.
    Last edited by Plato; April 22nd 2007 at 01:21 PM. Reason: Post Script
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