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How to find the derivative with an absolute value?

  1. Oct 17, 2012 #1
    I need to find the derivative of y=| x | / squareroot of 2-x2.

    We never learned how to find the derivative with an absolute value, so I have absolutely no idea how to do this problem and I can't find an example in the textbook.
     
  2. jcsd
  3. Oct 17, 2012 #2

    SammyS

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    Express this as a piecewise defined function.
     
  4. Oct 17, 2012 #3

    Ray Vickson

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    Look at the two cases x > 0 and x < 0.

    RGV
     
  5. Oct 17, 2012 #4
    I ended up with the derivatives:

    x2/[(2-x2)(squareroot of 2-x2)] + 1/(squareroot of 2-x2)

    -x2/[(2-x2)(squareroot of 2-x2)] - 1/(squareroot of 2-x2)


    Now, I need to find the tangent line to the curve at (1,1). I'm following an example from the textbook, but I keep ending up with y=2x-1, which passes through the original curve twice. Here's what I did:

    1.) I substituted 1 for x in the derivative and got y=2, which I used as the slope.
    2.) Then I used slope intercept form to solve for b. I had x=1, y=1, and m=2.
    3.) So, I end up with the equation y=2x-1.
     
  6. Oct 17, 2012 #5
    Where do you go to school? I'm taking intro to analysis and i've never taken the derivative of the absolute value. I wish I did, your school seems a bit tough. What book do you use? Perhaps I can download it and help you more.
     
  7. Oct 17, 2012 #6
    No, you don't want to go to my school. It sucks and I'm transferring next year. And my textbook is Calculus 7E Early Transcendentals by James Stewart. The question is on Chapter 3.4 #56. And, apparently, it can be downloaded online. I wish I had known that before buying my book.
     
  8. Oct 17, 2012 #7
    Neat i have that book, let me check it out. That's a solid calculus book there. My calculus teacher uses it, he likes it a lot.
     
  9. Oct 17, 2012 #8

    HallsofIvy

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    I assume these are for different values of x. You say what the values of x are. That is an important part of the answer.

    Since x is positive in small neighborhoods of 1, You can simply ignore the absolute value.
    You mean you got y'= 2 at x= 1.

     
  10. Oct 17, 2012 #9
    You've got to check out the graph of the function. The absolute value doesn't matter because if you check out the graph, and probably because it's a function and you're trying to find a tangent at any point x, you don't need the absolute value, so let me try to solve that problem here...

    I think it's y = 2x +1

    Just check your point algebra and then graph both functions on your calculator to check.
     
  11. Oct 17, 2012 #10
    ^your algebra, I was tying point slope form and I'm falling asleep here.
     
  12. Oct 17, 2012 #11
    ^ trying to type point slope form, sorry.
     
  13. Oct 17, 2012 #12
    Well, I feel dumb. I was graphing (x / squareroot of 2-x2) and (-x / squareroot of 2-x2) separately. I was getting two s-shaped graphs. I could have been finished with this question a long time ago. Well, thanks for your help.
     
  14. Oct 17, 2012 #13

    Ray Vickson

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    You definitely DO need the absolute value. The two functions f1(x) = |x|/sqrt(2-x^2) and f2(x) = x/sqrt(2-x^2) have very different graphs (although the two graphs coincide in the region x > 0). Furthermore, f2(x) is continuously differentiable in the region (-1/sqrt(2) < x < 1/sqrt(2)), while f1(x) is not (it does not have a derivative at x = 0).

    RGV
     
  15. Oct 18, 2012 #14
    I don't follow your reasoning, I'm definitely not saying I'm right, but if it's a function, then at each x value there is a unique output, so on it's graph, at x=1 it has one y coordinate.

    Where my reasoning could certainly be flawed is in this, I assumed that the positive region of the absolute value corresponded to positive x values. If this is true, and I do not know, then what I said seems right to me.
     
  16. Oct 18, 2012 #15
    Wait yeah you are taking this out of the context of the problem, the problem in the book asks you to simply find the tangent line at (1,1), that's all. As for the theory, it's beyond my understanding and the question he or she is asking.
     
  17. Oct 18, 2012 #16

    Ray Vickson

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    Well, he/she first asked how to take the derivative of |x|/sqrt(2-x^2). In a LATER posting, he/she said "Now, I need to find the tangent line to the curve at (1,1)". Of course the tangent line at x = 1 only needs the derivative in the region {x > 0}, but that is not what the original problem said. Maybe the OP did not actually post the precise problem, but I can only go by what is written.

    RGV
     
  18. Oct 18, 2012 #17
    Right, I looked in the textbook, it just says find the tangent line of the original function at (1,1), it's an introductory calculus problem. I don't even understand your analysis of it, but I respect it.
     
  19. Oct 18, 2012 #18

    Ray Vickson

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    I did not analyse anything; I just suggested breaking up the problem into two parts (x > 0 or x < 0). After all, |x| is defined differently in those two parts, so it makes sense to do that.

    RGV
     
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