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Math Help - Second derivative help

  1. #1
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    Second derivative help

    Just wanted to see if I'm doing this correctly:

    Find f''(x)=|x|^3

    So f'(x)=(3|x|^2)\frac{x}{|x|}

    =3|x|x

    f''(x)=3|x|\frac{d}{dx}x+3\frac{d}{dx}|x|x

    =3|x|+3\frac{x}{|x|}x

    =3|x|+\frac{3x^2}{|x|}
    Does that seem right?
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  2. #2
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    Quote Originally Posted by ascendancy523 View Post
    Just wanted to see if I'm doing this correctly:

    Find f''(x)=|x|^3

    So f'(x)=(3|x|^2)\frac{x}{|x|}

    =3|x|x

    f''(x)=3|x|\frac{d}{dx}x+3\frac{d}{dx}|x|x

    =3|x|+3\frac{x}{|x|}x

    =3|x|+\frac{3x^2}{|x|}
    Does that seem right?

    First of all your first line is your second derivative... you need to ask : What is the second derivative of f(x)

    Hint:

    f(x)=|x|^3=sgn(x)\cdot x^3

    Now, is much better looking...

    You can see straight that:

    f''(x)=6\sqrt{x^2}
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  3. #3
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    Quote Originally Posted by ascendancy523 View Post
    Just wanted to see if I'm doing this correctly:

    Find f''(x)=|x|^3
    This is written badly. What this says is that f''(x) is |x|^3. What you mean is "Find f''(x) where f(x)= |x|^3

    So f'(x)=(3|x|^2)\frac{x}{|x|}

    =3|x|x

    f''(x)=3|x|\frac{d}{dx}x+3\frac{d}{dx}|x|x

    =3|x|+3\frac{x}{|x|}x

    =3|x|+\frac{3x^2}{|x|}
    Does that seem right?
    Of course, x^2= |x|^2
    so this is the same as 3|x|+ 3\frac{|x|^2}{|x|}= 6|x| for x\ne 0 and is not defined for x= 0.

    I would consider it simpler (and better) to say "If x\ge 0 f(x)= x^3 so f'(x)= 3x^2 and f"(x)= 6x, if x< 0,  f(x)= -x^3 so f'(x)= -3x^2 and f''(x)= -6x. Of course, |x| is not differentiable at x= 0 so neither is |x|^3 and so f'' does not exist at x= 0".
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    This is written badly. What this says is that f''(x) is |x|^3. What you mean is "Find f''(x) where f(x)= |x|^3


    Of course, x^2= |x|^2
    so this is the same as 3|x|+ 3\frac{|x|^2}{|x|}= 6|x| for x\ne 0 and is not defined for x= 0.

    I would consider it simpler (and better) to say "If x\ge 0 f(x)= x^3 so f'(x)= 3x^2 and f"(x)= 6x, if x< 0,  f(x)= -x^3 so f'(x)= -3x^2 and f''(x)= -6x. Of course, |x| is not differentiable at x= 0 so neither is |x|^3 and so f'' does not exist at x= 0".
    Yeah that would make it much simpler. Thanks for the help!
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