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We have to differentiate this function at $x=0$ $$\left. \dfrac{d \arcsin (\cos (x))}{dx} \right|_{x=0}$$ Using the identity $\cos x=\sin \Big(\dfrac{\pi}{2}-x\Big)$ we get $$\left. \dfrac{d \arcsin (\sin (\dfrac{\pi}{2} - x))}{dx} \right|_{x=0},$$ For $0 \le x \le \pi$. We know $\arcsin \Big(\sin \Big(\dfrac{\pi}{2}-x\Big)\Big)=\dfrac{\pi}{2}-x$.

So, our original problem reduces to $$\left. \dfrac{d\Big(\dfrac{\pi}{2}-x\Big)}{dx}\right|_{x=0},$$ which is equal to $-1$ for $0 \le x \le \pi$.

But the derivative of this function at $x=0$ is undefined. What's going on here?

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  • $\begingroup$ I don't know what's going on for I didn't look at the problem carefully, but don't you worry about the case $x<0$? $\endgroup$ – Shashi Jul 11 '18 at 11:02
  • $\begingroup$ @Adam It generally does. At least for the inequality I mentioned above. $\endgroup$ – user571036 Jul 11 '18 at 11:20
  • $\begingroup$ @Shashi Considering the case at x=0 gave me enough trouble. Firstly, help me to get out of that. Then, I will consider x being lesser than 0. $\endgroup$ – user571036 Jul 11 '18 at 11:22
  • $\begingroup$ something to do with arcsin(sin(pi/2 - x)) = pi/2 - x the function can generate different values $\endgroup$ – Cato Jul 11 '18 at 11:29
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    $\begingroup$ What Shashi wanted to say is that $\arcsin(sin(\pi/2-x))$ isn't always $\pi/2-x$ : $\arcsin(sin\theta))=\theta$ only if $-\pi/2\le\theta\le\pi/2$, but your $\theta=\pi/2-x$ doesn't belong to this interval for $x<0$ (which you have to consider for you derivative). $\endgroup$ – Nicolas FRANCOIS Jul 11 '18 at 11:29
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Near zero you have $$\arcsin(\cos x) = \frac{\pi}{2} - \arccos(\cos x) = \frac{\pi}{2} - |x|$$ (so the derivative is $+1$ for $x<0$ and $-1$ for $x>0$). And therefore the function cannot be approximated by a linear function in a arbitrary small interval containg $0,$ and this implies that the derivative at $x=0$ does not exist.

Edit: Near $x = 0$ you have $\arccos(\cos x)=|x|$ because $\cos(x)$ is an even function. It cannot be just $x$ because the value is positive for $x\ne0$.

You also have to look at values $x<0,$ if you want to know whether the function has a derivative in a neighborhood of $0$ (it has, if the left and rights derivatives exist and are equal, see e.g. here).

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  • $\begingroup$ Is $\arccos (cosx)$ equal to x or, |x| ? $\endgroup$ – user571036 Jul 11 '18 at 11:51
  • $\begingroup$ @user571036: See edit. $\endgroup$ – gammatester Jul 11 '18 at 12:04
  • $\begingroup$ @user571036 Yes, you are right.Derivative of Arcsin(cos(x)) is undefined @x=0 Its limit from the left is $\lim_{x\to0^-}\frac{\partial(\arcsin\cos(x))}{\partial{x}}=1$. Its limit from the right is $\lim_{x\to0^+}\frac{\partial(\arcsin\cos(x))}{\partial{x}}=-1$ $\endgroup$ – Dhamnekar Winod Jul 11 '18 at 13:03
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The derivative of the function, for $x\ne0$, is $$ -\frac{\sin x}{\sqrt{1-\cos^2x}}=-\frac{\sin x}{\lvert\sin x\rvert} $$ By l'Hôpital, $$ \lim_{x\to0^+}\frac{\arcsin(\cos x)-\arcsin(\cos0)}{x}= \lim_{x\to0^+}-\frac{\sin x}{\lvert\sin x\rvert}=-1 $$ and, similarly, the limit from the left is $1$.

Thus the limit of the difference quotient doesn't exist.


If we limit the function to the interval $(-\pi/2,\pi/2)$, we have that $$ f'(x)=\begin{cases} 1 & -\pi/2<x<0 \\[4px] -1 & 0<x<\pi/2 \end{cases} $$ so that $$ f(x)=\begin{cases} x+c_- & -\pi/2<x<0 \\[4px] -x+c_+ & 0<x<\pi/2 \end{cases} $$ for some constants $c_-$ and $c_+$; continuity at $0$ and $f(0)=\pi/2$ gives $c_-=c_+\pi/2$.

Now it should be clear what's your mistake: the equality $$ \arcsin(\sin(\pi/2-x)=\pi/2-x $$ doesn't hold for all $x$.

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