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Thread: Proving x|x| is differentiable

  1. #1
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    Proving x|x| is differentiable

    I just have a quick, and most likely silly question. I need to show that the function x|--> x|x| is differentiable on R. Would the best way to do this be to let g(x) = x, h(x) = |x|, and use the product rule to show that the derivative is 2x for all x?

    Is there anything special I would need to do for x = 0, since |x| isn't differentiable at x=0?

    Thanks for your help.
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  2. #2
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    Quote Originally Posted by monsieur fatso View Post
    I just have a quick, and most likely silly question. I need to show that the function x|--> x|x| is differentiable on R. Would the best way to do this be to let g(x) = x, h(x) = |x|, and use the product rule to show that the derivative is 2x for all x?

    Is there anything special I would need to do for x = 0, since |x| isn't differentiable at x=0?

    Thanks for your help.
    $\displaystyle f(x) = x|x|$

    for the case $\displaystyle x = 0$ ...

    $\displaystyle f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}$

    $\displaystyle f'(0) = \lim_{x \to 0} \frac{x|x|}{x} = 0$

    should be clear that the limit also exists for all $\displaystyle x \neq 0$
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    Okay, I guess I'm having some trouble evaluating that limit..
    $\displaystyle f'(x) = \lim_{t\to x} \frac{f(t)-f(x)}{t-x} = \frac{x|x|-x|x|}{x-x}=|x|
    $

    Which isn't correct...
    Any advice?
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  4. #4
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    $\displaystyle f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

    $\displaystyle f'(a) = \lim_{x \to a} \frac{x|x| - a|a|}{x - a}$

    for $\displaystyle x$ and $\displaystyle a$ both positive ...

    $\displaystyle f'(a) = \lim_{x \to a} \frac{x^2 - a^2}{x - a}
    $

    $\displaystyle f'(a) = \lim_{x \to a} (x + a) = 2a$

    for $\displaystyle x$ and $\displaystyle a$ both negative ...

    $\displaystyle f'(a) = \lim_{x \to a} \frac{x(-x) - a(-a)}{x - a}$

    $\displaystyle f'(a) = -\lim_{x \to a} \frac{x^2 - a^2}{x - a}$

    $\displaystyle f'(a) = -\lim_{x \ to a} (x + a) = -2a
    $
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