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Differentiablity of sinx at pi/2

  1. May 20, 2010 #1
    Is f(x)=sin x differentiable at x = pi/2. If yes why?

    because what i think is at pi/2 f'(x) changes from 1 to -1 so at x=pi/2 it may not be differentiable (similar to case of |x|).
     
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  3. May 20, 2010 #2

    Cyosis

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    Why do you think f'(x) suddenly changes from 1 to -1 at pi/2?
     
  4. May 20, 2010 #3
    zoom in to the area where it changes sign ..... there cerrtainly will be a region where sign will get changed .......it's obvious
     
  5. May 20, 2010 #4

    Cyosis

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    There is an extremely major difference between gradually changing sign and a jump from 1 to -1 at pi/2 (as you claim). My question is why do you think it goes from 1 to -1 at x=pi/2?
     
  6. May 20, 2010 #5
    i agree that it gradually changes sign (since the graph is curved and not straight as in |x|) . But in LHS and RHS of pi/2 slopes will be opposite ,not -1 and 1 but they will be opposite to each other (due to symmetry of curve) , so pi/2 is a point where slope changes from LHS to RHS. Thus it shouldn't be diferentiable at that point....
     
  7. May 20, 2010 #6

    Cyosis

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    The signs will be opposite to each other yes, but they won't be -1 and 1 except for two points. Intuitively, if you can draw a unique tangent line to the curve at a point then the function has a derivative at that point. Can you draw a tangent line at the point x=Pi/2 for sin x? Can you draw one for the function |x| at x=0?
     
  8. May 20, 2010 #7
    i understand what you are trying to say but i am still confused.......

    See, i'll tell you where i'm confused....
    A function is differentiable at point xo if it the derivative is continous as xo . Now derivative will be continous at xo if it exists at neighbourhood points say x1 and x2 and is equal to it's value at xo. Right? Now consider point x1 (at Lhs of xo)... at Lhs of x1 say at x3 sllope will be same as that of x1 and xo(at RHS o x1) for derivative to be continous at x1......now consider point x3 and repeat the same procedure and then x4 and u'll see that slope remains constant..........
     
  9. May 20, 2010 #8

    HallsofIvy

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    This is certainly NOT true. A function is differentiable at x0 if and only if its derivative exists- it is not necessary that the derivative be continuous there. A function is said to be "continuously differentible" at a point if the derivative is continuous there, as stronger condition than just being differentiable. And, in fact, sin(x) is "continuously differentiable" at [itex]\pi/2[/itex].

    No, not right! You seem to be saying that only constant functions are continuous! A function is continuous at x= a if the limit, as x goes to a is f(a). That, in turn, means that, given any [itex]\epsilon> 0[/itex], there exist a neighborhood in which [itex]|f(x)- f(a)|< \epsilon[/itex]. It is certainly not necessary that any points in that neighborhood, other than a, have the same value as at a.

    Your are stating things about continuity and differentiability that are simply not true. You need to review both concepts.

    The derivative of sin(x) is cos(x) and [itex]cos(\pi/2)= 0[/itex]. That's how the derivative can change sign continuously at [itex]\pi/2[/itex].
     
  10. May 20, 2010 #9
    can u give me example of a function whose derivative at particular point exists but the derivative is not continous at that point?
    When did i say only contant functions are continous. By stating points x1 and x2 i meant these points are infinitely close to xo at LHL and RHL, repectively. Lt x->a
    means x comes very very closer to a. And that's what i meant!
    Moreover, a function is continous when LHL = F(a) = RHL.
     
  11. May 20, 2010 #10
    The example you gave in the first post is such a function: |x|.
    That function is perfectly differentiable (it will be 1 or -1) throughout the domain, but it changes sign with 'infinite' slope at x=0.
     
  12. May 20, 2010 #11
    [tex]
    f(x) = x^2 sin\frac{1}{x}
    [/tex]

    Where f(0) = 0. Clearly, f is continuous at zero, as you can see by taking a limit. It is also differentiable at zero, as you can see by doing the definiton.
    [tex]
    \lim_{h \rightarrow 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \rightarrow 0} \frac{f(h) - 0}{h}
    [/tex]
    [tex]
    \lim_{h \rightarrow 0} \frac{h^2 sin\frac{1}{h}}{h} = \lim_{h \rightarrow 0} h sin \frac{1}{h} = 0
    [/tex]

    So [tex] f'(0) = 0[/tex]

    But...
    [tex]
    f'(x) = 2x sin \frac{1}{x} - cos\frac{1}{x}
    [/tex]
    Which is obviously discontinuous at x = 0.
     
  13. May 21, 2010 #12

    HallsofIvy

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    No. |x| is NOT differentiable at x= 0.

    The function L'Hopital gives, above, [tex]x^2 sin(\frac{1}{x})[/tex], is differentiable for all x but the derivative is not continuous at x= 0.
     
    Last edited by a moderator: May 22, 2010
  14. May 21, 2010 #13

    HallsofIvy

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    L'Hopital gave an example above: [tex]x^2 sin(\frac{1}{x})[/tex] is continuous for all x butt the derivative is not continuous at x= 0.

    You said "Now derivative will be continous at xo if it exists at neighbourhood points say x1 and x2 and is equal to it's value at xo." If, for any neighborhood, there exist points x1 and x2 such that f(x1)= f(x2)= f(x0), then f must be constant near x0.

    There are no such thing as "points infinitely close to x0". You have a very strange notion of what a limit is. "Let x-> a" does NOT mean "x comes very very close to a". You seem to be trying to talk about infinitesmals, a very deep and abstract way to talk about limits, but I suspect that you are not clear on what they are.
     
  15. May 21, 2010 #14

    Char. Limit

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    Was it necessary to post the same content three times?

    sin(x) is differentiable for all x because cos(x) exists for all x. Simple.
     
  16. May 21, 2010 #15
    ^ It was just a lag issue. That may happen when you post a reply, the page starts to load, you input a few more words, you click it again, and etc.

    If it's obvious that it's changing sign gradually can't you see that the limits of both the right hand and the left hand of f'(x) would reach 0? That is, the slope approaches 0 as x reaches pi/2.
     
  17. May 22, 2010 #16
    OK now i understand i think!
    In function sinx slope gradually changes so at pi/2 we have only one tangent whose slope is 0.
    Whereas in functions like |x| at x=0 infinite tangents in between -1 and 1 are possible .Hence derivative is not defined at x=0.
    Am i right now?
     
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