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Math Help - Chain Rule

  1. #1
    Member mybrohshi5's Avatar
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    Chain Rule

    Write absolute value(x)=sqrt(x^2) and use the chain rule to show that

    (d/dx)abs(x)=x/abs(x)

    could i get some help with this please?

    i know the chain rule is F'(x)=f'(g(x))*g'(x) but i do not know how to apply that to this problem.

    thank you
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  2. #2
    Super Member Matt Westwood's Avatar
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    First you need to divide the domain into two bits to get rid of that pesky "absolute value" bit.

    When x > 0, abs x = x.
    When x < 0, abs x = -x.

    So first differentiate f(x) = +\sqrt {x^2} for positive x and then do the same but minus it for negative x. You'll get two expressions that you can combine by using the abs function as suggested in the answer you're directed towards.
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  3. #3
    Member mybrohshi5's Avatar
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    i kinda get what you are saying but i guess im just kind of confused because isnt the sqrt(x^2) just x? and the derivative of x is just 1?
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  4. #4
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by mybrohshi5 View Post
    i kinda get what you are saying but i guess im just kind of confused because isnt the sqrt(x^2) just x? and the derivative of x is just 1?
    Yep, that's right. Except that the derivative for x < 1 is -1 (you have to take the negative of it, remember?).

    So: what's abs x when x > 1? It's x.

    So what's abs x / x when x > 1? It's x / x = 1.

    Now: what's abs x when x < 1? It's -x.

    So what's abs x / x when x < 1? It's \frac {-x} x which is .... ?
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  5. #5
    Member mybrohshi5's Avatar
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    it is -1,
    but i dont get how this is helping to show that (d/dx)abs(x)=x/abs(x)

    i mean i understand what you are doing for the most part but im not getting how to put this all together to help show that (d/dx)abs(x)=x/abs(x)

    could you please explain

    thank you very much =)
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  6. #6
    Super Member Matt Westwood's Avatar
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    You want to find d/dx(abs x).

    So you use the identity that abs x = +\sqrt {x^2}.

    So you differentiate \sqrt {x^2} using the chain rule, noticing that it's 1 for x > 0 and it's -1 for x < 0.

    So, oh good, we know the driv of \sqrt {x^2}.

    But this is the same as the driv of abs x.

    So d/dx(abs x) equals 1 for x > 0 and -1 for x < 0.

    That's the main result.

    The final bit is showing that abs x / x is *also* equal to 1 when x > 0 and -1 when x < 0.

    Equate the two.

    Bingo. You've now proved what you were asked to.
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  7. #7
    Super Member Matt Westwood's Avatar
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    If you still have problems, get someone else to help because it's nearly 1:30 and I'm going off line.
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  8. #8
    Member mybrohshi5's Avatar
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    is it because the f'(x) of +sqrt(x^2) is 1 and f'(x) of -sqrt(x^2) is -1 so....

    (d/dx)abs(x) for x>1 will be +1 and for (d/dx)abs(x) for x<1 will be -1?
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  9. #9
    Member mybrohshi5's Avatar
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    thank you =) i get it all now
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