0
$\begingroup$

according to me arcsin should not be differentiable at $x=1$ as it has infinite slope at that point. But the books say otherwise...

$ y=\arcsin x $

$dy/dx= 1/\sqrt(1-x^2) $

at $x=1$ it is not defined... pls explain..

  • 1
    arcsin(1) is very defined...it is equal to pi/2 – 2017-01-27

3 Answers 3

1

You can solve this using the definition of the derivative: Take $L = \lim_{x\to_1}\frac{\arcsin x - \arcsin 1}{x-1} = \lim_{x\to1}\frac{\arcsin x - \frac{\pi}{2}}{x-1}$. Now let $x = \sin u$ (observe that $u\to\frac{\pi}{2}$), then you get $L = \lim_{u\to\frac{\pi}{2}}\frac{u- \frac{\pi}{2}}{\sin u-1}$. Using l'Hôpital's rule, we get $L = \lim_{u\to\frac{\pi}{2}}\frac{1}{\cos u} = \frac{1}{0} = \infty$. So the derivative is not defined in $x=1$.

Note: This is the limit for $x\to1$ with $x<1$, because $\arcsin$ is not defined for $x>1$.

0

As you can see, $1/(\sqrt{1-x^2})$ is not defined on $x=1$

The expression $1/(\sqrt{1-x^2})$ is the derivative of $\arcsin(x)$ for those regions where the function is derivable. Since it has a discontinuity on $x = \pm 1$, it is not derivable there. The domain of this function is $(-1,1)$.

  • 0
    according to me arcsin is not discontinuous anywhere... – 2017-01-27
  • 0
    arcsin is left-continuous but right-discontinuous for x = 1, since it isn't defined for x>1. – 2017-01-27
  • 1
    i dont think that is called discontinuity...discontinuity is when we have to lift our pencil while drawing the curve... – 2017-01-27
  • 0
    Discontinuity in some point $a$ for a function f(x) means that the $\lim_{x\to a} ≠ f(a)$, the limit is infinity or that the limit is not defined at all ($f(x)$ is continuous in $a$ if and only if $\lim_{x\to a} = f(a)$). An intuitive way of understanding that is indeed that you have to lift your pencil when drawing the curve. Note that arcsin is continuous in the open interval ]-1, 1[, right-continuous in -1 and left-continuous in 1. For all other values of x is arcsin x discontinuous – 2017-01-27
0

Let's see it geometrically. As you know $sin^{-1}x$ is only defined in [-1,1], so $(-1,-\pi/2)$ and $(1,\pi/2)$ are the two end points of the graph. So, you see we cannot talk about tangent to the end points of a curve. We cannot draw any unique line which touches the curve of arcsinx at the end points, and hence the slope of the curve is not defined there. I think that maybe the left hand derivative os defined there. The left hand derivative should be zero there (I think) but it doesn't make sense to talk about approaching the curve from right because the curve doesn't exist to the right of x=1.

  • 0
    acc. to what ive learnt, if a function isnt defined to the right, we just check the left side and the point itself... – 2017-01-27
  • 0
    The definition of derivative, $\lim_{y\to x}\frac{f(y)-f(x)}{y-x}$ makes sense at an end point if the definition of limit makes sense at an end point at all. I would say it does, but that's almost a matter of taste. – 2017-01-27
  • 0
    @Arthur: I don't think that the definition of limit makes any sense at an end point. A limit exists only if the left hand and right hand limits are equal. If a function isn't defined to the right of a particular point, then the right hand limit and hence the limit doesn't exist. – 2017-01-29
  • 0
    @Aishit Dharwal: From what I've learnt, for the derivatibe to exist, both the left hand and right hand derivatibes should exist and should be equal. And, we don't need to check the value of the function at that point for checking differentiability. – 2017-01-29