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Math Help - prove that if f and g are differentiable at a then fg is differentiable at a

  1. #1
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    prove that if f and g are differentiable at a then fg is differentiable at a

    Prove that if f and g are differentiable at a then fg is differentiable at a.

    I'm not sure how to start this. I said that fg'(a) =lim(x->a) ((f(x)-f(a))(g(x)-g(a))/(x-a)^2. Is this correct, if so how do I proceed?
    Last edited by mr fantastic; May 14th 2011 at 05:48 AM. Reason: Copied title into main body of post.
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  2. #2
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    I assume you mean using the basic definition of the derivative rather than the product rule.

    No, what you have written is NOT correct. The derivative of a product is NOT the product of the derivatives. The derivative would be given by
    \lim_{x\to a}\frac{f(x)g(x)- f(a)g(a)}{x- a}

    The crucial point you need is that f(x)g(x)- f(a)g(a)= f(x)g(x)- f(a)g(x)+ f(a)g(x)- f(a)g(a)= (f(x)- f(a))g(x)+ f(a)(g(x)- g(a)).
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  3. #3
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    Quote Originally Posted by poirot View Post
    I'm not sure how to start this. I said that fg'(a) =lim(x->a) ((f(x)-f(a))(g(x)-g(a))/(x-a)^2. Is this correct, if so how do I proceed?
    No that is not correct. You want
    \lim _{h \to 0} \frac{{f(a + h)g(a + h) - f(a)g(a)}}{h}.
    Add and subtract f(a)g(a+h) in the numerator.
    Because g is continuous at a we know that \lim _{h \to 0} g(a + h) = g(a).
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    Sorry I don't understand your post. How did you begin and why would add something, then subtract it?
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  5. #5
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    Quote Originally Posted by poirot View Post
    Sorry I don't understand your post. How did you begin and why would add something, then subtract it?
    You are questioning what is actually a standard approach in many calculus textbooks for deriving the product rule.

    I suggest you Google the derivation or go to your institute's library and find a Calculus textbook that contains the derivation (eg. Sallas and Hille or Thomas and Finney).
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    Quote Originally Posted by poirot View Post
    Sorry I don't understand your post. How did you begin and why would add something, then subtract it?
    Look at reply #2.
    \frac{{f(x)g(x) - f(a)g(x) + f(a)g(x) - f(a)g(a)}}{{x - a}}

    =\frac{{f(x) - f(a)}}{{x - a}}g(x) + f(a)\frac{{g(x) - g(a)}}{{x - a}}.
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    Ok, so as x tend to a g(x) tends to g(a) and so we get g(a)f'(a)+f(a)g'(a). I.e the familiar product rule. One question, at the begining it is as though fg(x) =f(x)g(x) in the numerator. Is this using the algebra of limits, Limit of fg =limit of f *limit of g? Thanks for your help.
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  8. #8
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    Quote Originally Posted by poirot View Post
    One question, at the begining it is as though fg(x) =f(x)g(x) in the numerator. Is this using the algebra of limits, Limit of fg =limit of f *limit of g? Thanks for your help.
    I am not fully clear as to the meaning of the question.
    Is it about notation?
    For functions f~\&~g the notation fg(a) means f(a)g(a)~.
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  9. #9
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    Yes sorry I misunderstood the notation.
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