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Math Help - Arcsinx

  1. #1
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    Arcsinx

    So there is a question that asks that to show that sinx has an inverse function, and show that it;s differantiable at any point. So I showed that it's one to one and it's bijection func at [-pie/2,pie/2], and ther3fore exists an inverse function in that interval. However I am having a trouble showing that it's differantiable. I had an idea to show that 1/cosx is defined in (-pie/2,pie/2) and therefore according to the inverse derivative it's differantiable at any point in that interval if there is a derivative of that function. However, I think there is a problem with my logic, because according to what I did ,I think , I already assumed that it's differantiable.
    The other thing that I did , I showed that the inverse function is continuous at all points and it's an increasing function. So, in my point of view, every function that is increasing and it's continuous , is differantiable. However I couldn't prove that.
    I tried to play with the definition of differentiation and somehow get something . however I failed. Can someone give me a hint. Please note that , at this point they assume that we know nothing about arcsin. so if there is a trigonometric identity, I cant use that.
    Thank you.
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    Re: Arcsinx

    From dy=f'(x)dx, you can conclude dx=(1/f'(x))dy. Look it up. I don't remember the details.
    Thanks from davidciprut
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    Re: Arcsinx

    Quote Originally Posted by davidciprut View Post
    So there is a question that asks that to show that sinx has an inverse function, and show that it;s differantiable at any point. So I showed that it's one to one and it's bijection func at [-pi/2,pi/2], and ther3fore exists an inverse function in that interval. However I am having a trouble showing that it's differantiable. I had an idea to show that 1/cosx is defined in (-pi/2,pi/2) and therefore according to the inverse derivative it's differantiable at any point in that interval if there is a derivative of that function. However, I think there is a problem with my logic, because according to what I did ,I think , I already assumed that it's differantiable.
    There are many ways of doing this. Here is only one.
    If $y=\sin(x)$ then think of $x$ as the measure of an angle in a right triangle with one side along the positive x-axis with length $\sqrt{1-y^2}$.

    This means that $-\dfrac{\pi}{2}\le x\le \dfrac{\pi}{2}$ and $-1\le y\le 1$.

    The define $\displaystyle\arcsin(y)=\int_0^y {\frac{{dt}}{{\sqrt {1 - {t^2}} }}} $
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    Re: Arcsinx

    To show sinx is differentiable, use definition of derivative. See any calculus text.
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    Re: Arcsinx

    I can't use the facts that you guys have provided because, although I know everyone of them, technically we don't know it yet. It's a question in my homework but I am very limited that's why I am stuck for hours, I have a proof, but I don't know if it's correct I would be happy if you guys can verify it. I used it based on the theorem we learned on inverse func derivatives.

    By the way, is this prepositon correct?
    Preposition:f is defined on closed interval [a,b] and continuous. If f is increasing then f is differentiable.

    I think if I can prove this I can prove it too, I tried, however I failed, can someone tell me if it's true or not ? if it's not it's not worth wasting time on. I think it's correct because the derivative is bounded, and it prevents the limit (of derivative definition) going to infinity, or it prevents of having a point like x=0 in |x| func.
    Attached Thumbnails Attached Thumbnails Arcsinx-arcsiny.png  
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    Re: Arcsinx

    By the way , the credit goes to you guys because I got the idea from both of you
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    Re: Arcsinx

    From:
    Inverse functions and differentiation - Wikipedia, the free encyclopedia
    "Writing explicitly the dependence of y on x and the point at which the differentiation takes place and using Lagrange's notation, the formula for the derivative of the inverse becomes

    [f-1]'(a) = 1/f'(f-1)(a))*

    Geometrically, a function and inverse function have graphs that are reflections, in the line y = x. This reflection operation turns the gradient of any line into its reciprocal.
    Assuming that f has an inverse in a neighbourhood of x and that its derivative at that point is non-zero, its inverse is guaranteed to be differentiable at x and have a derivative given by the above formula."

    Referring to the same formula:
    Derivatives of Inverse Functions
    observes:
    “We shall not discuss the proof of this theorem here, other than to say that the proof is relatively difficult. A first step in the proof is to show that the inverse of a continuous function is continuous; this proof in turn requires an application of the Intermediate Value Theorem (see Stage 4) and actually reveals a fairly deep and subtle property of the real numbers”

    * same as formula I gave above.

    In general, google "derivative of an inverse"

    As usual, I find Plato's post unintelligible in regard to this thread.
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    Re: Arcsinx

    Quote Originally Posted by Hartlw View Post
    As usual, I find Plato's post unintelligible in regard to this thread.
    That is because you know absolutely nothing about the topic in this thread.
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    Re: Arcsinx

    Quote Originally Posted by Plato View Post
    There are many ways of doing this. Here is only one.
    If $y=\sin(x)$ then think of $x$ as the measure of an angle in a right triangle with one side along the positive x-axis with length $\sqrt{1-y^2}$.

    This means that $-\dfrac{\pi}{2}\le x\le \dfrac{\pi}{2}$ and $-1\le y\le 1$.

    The define $\displaystyle\arcsin(y)=\int_0^y {\frac{{dt}}{{\sqrt {1 - {t^2}} }}} $
    Your formula is just the mindless inverse of the formula for the derivative of arcsinx from any table of derivatives.

    If your formula defines arcsinx, what is sinx using your definition for arcsinx.

    If you are going to reference a picture to define x, then arcsinx is continuous from the picture.
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    Re: Arcsinx

    Quote Originally Posted by Hartlw View Post
    Your formula is just the mindless inverse of the formula for the derivative of arcsinx from any table of derivatives. If your formula defines arcsinx, what is sinx using your definition for arcsinx. If you are going to reference a picture to define x, then arcsinx is continuous from the picture.
    The above response shows how ignorant you really are about the foundations of mathematics.
    I actually quoted the definition given by Leonard Gillman from my favorite calculus textbook.
    Now before you complain that this is obscure (you have done this out of your profound ignorance in the past) please look at this.
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    Re: Arcsinx

    From elementary calculus:
    Definition: θ=arclength along unit circle from point (1,0) to point (x,y)
    Definition: cosθ=x, sinθ=y
    It is then shown that sinx is continuous, and dsinx/dx=cosx.

    Switching notation, we then have siny is continuous and it’s derivative exists for all y.
    It then follows (see my previous posts) that:
    x = siny → dx=cosydy → dy/dx = 1/cosx = 1/(1-x2)1/2
    So:
    y = arcsinx
    d(arcsinx)/dx = 1/(1-x2)1/2

    ∫1/(1-x2)1/2=arcsinx by definition of integral.

    There are endless abstractions of fundamental principles, which are faily easy to come up with with a little imagination.

    The fact that you happen to have come across one which someone else hasn’t doesn’t make you intelligent or them ignorant.

    You could, without any explanation or reference, recite Kelley’s definition of uniform continuity and then call me ignorant because I didn’t recognize it or understand it. That would’nt make you intelligent or me ignorant, it would just make you devious.

    EDIT: Of course inverse function has a domain of definition over which it is single valued.
    Last edited by Hartlw; February 26th 2014 at 04:55 AM.
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