How do you differentiate arcsin(sqrt(sin^2(1/x) ) using the chain rule?

Apr 14, 2016

Initially, I said I would not use the chain rule. But, see below. (Warning, this looks more complicated to me than using the chain rule turned out to be.)

Explanation:

Use $\sqrt{{u}^{2}} = \left\mid u \right\mid = \left\{\begin{matrix}u & \text{if" & u >= 0 \\ -u & "if} & u < 0\end{matrix}\right.$, to get

$f \left(x\right) = \arcsin \left(\sqrt{{\sin}^{2} \left(\frac{1}{x}\right)}\right) = \arcsin \left(\left\mid \sin \left(\frac{1}{x}\right) \right\mid\right)$

$= \left\{\begin{matrix}\arcsin \left(\sin \left(\frac{1}{x}\right)\right) & \text{if" & sin(1/x) >= 0 \\ arcsin(-sin(1/x)) & "if} & \sin \left(\frac{1}{x}\right) < 0\end{matrix}\right.$

$= \left\{\begin{matrix}\arcsin \left(\sin \left(\frac{1}{x}\right)\right) & \text{if" & sin(1/x) >= 0 \\ arcsin(sin(-1/x)) & "if} & \sin \left(\frac{1}{x}\right) < 0\end{matrix}\right.$

Now we have to find $\arcsin \left(\sin \left(\frac{1}{x}\right)\right)$ and $\arcsin \left(\sin \left(- \frac{1}{x}\right)\right)$ for the two cases $\sin \left(\frac{1}{x}\right) \ge 0$ and $\sin \left(\frac{1}{x}\right) < 0$

Positive sine

For every real number $x$ with $\sin \left(\frac{1}{x}\right) \ge 0$, we know $\frac{1}{x}$ is in Quadrant I or II.
So, there is exactly one integer $k$ such that $\frac{1}{x}$ lies in exactly one of the intervals below.

If $\frac{1}{x} \in \left[2 \pi k , \frac{\pi}{2} + 2 \pi k\right)$, then $\arcsin \left(\sin \left(\frac{1}{x}\right)\right) = \frac{1}{x} - 2 \pi k$.

If $\frac{1}{x} \in \left[\frac{\pi}{2} + 2 \pi k , \pi + 2 \pi k\right)$, then $\arcsin \left(\sin \left(\frac{1}{x}\right)\right) = \pi + 2 \pi k - \frac{1}{x}$.

Negative sine

For every real number $x$ with $\sin \left(\frac{1}{x}\right) < 0$, we know $\frac{1}{x}$ is in Quadrant III or IV. [And $- \frac{1}{x}$ is in II or I.]
So, there is exactly one integer $k$ such that $\frac{1}{x}$ lies in exactly one of the intervals below.

If $\frac{1}{x} \in \left[- \pi + 2 \pi k , - \frac{\pi}{2} + 2 \pi k\right)$, then $- \frac{1}{x}$ is in Quadrant II.
And, $- \frac{1}{x} \in \left[\frac{\pi}{2} - 2 \pi k , \pi - 2 \pi k\right)$ $\arcsin \left(\sin \left(- \frac{1}{x}\right)\right) = \pi - 2 \pi k - \left(- \frac{1}{x}\right)$.

If $\frac{1}{x} \in \left[- \frac{\pi}{2} + 2 \pi k , 2 \pi k\right)$, then $- \frac{1}{x}$ is in Quadrant I.
And, $- \frac{1}{x} \in \left[- 2 \pi k , \frac{\pi}{2} - 2 \pi k\right)$ $\arcsin \left(\sin \left(- \frac{1}{x}\right)\right) = \left(- \frac{1}{x}\right) + 2 \pi k$.

Writing $f \left(x\right)$

f(x) = {(1/x-2pik,"if",1/x in [2pik,pi/2+2pik)), (pi+2pik-1/x,"if",1/x in [pi/2+2pik,pi+2pik)), (pi-2pik+1/x,"if",1/x in [-pi+2pik,-pi/2+2pik)), (-1/x+2pik,"if",1/x in [-pi/2+2pik, 2pik)) :}

Differentiate each branch
We'll delete the joints, as the derivative will fail to exist if $\sin \left(\frac{1}{x}\right) = 0 \text{ or } 1$

f(x) = {(-1/x^2,"if",1/x in (2pik,pi/2+2pik)), (1/x^2,"if",1/x in (pi/2+2pik,pi+2pik)), (-1/x^2,"if",1/x in (-pi+2pik,-pi/2+2pik)), (1/x^2,"if",1/x in (-pi/2+2pik, 2pik)) :}

We might try to clarify by writing

f(x) = {(-1/x^2,"if",1/x" is in Quadrant I"), (1/x^2,"if",1/x" is in Quadrant II"), (-1/x^2,"if",1/x" is in Quadrant III"), (1/x^2,"if",1/x" is in Quadrat IV") :}

Finally observe that at the joints, the left and right derivatives are not equal, so the derivative does not exist at those points.

Apr 16, 2016

Using the chain rule, I get $\frac{- \sin \left(\frac{1}{x}\right) \cos \left(\frac{1}{x}\right)}{{x}^{2} \sqrt{{\sin}^{2} \left(\frac{1}{x}\right)} \sqrt{1 - {\sin}^{2} \left(\frac{1}{x}\right)}}$ which may be written: $- \frac{1}{x} ^ 2 \sin \frac{\frac{1}{x}}{\left\mid \sin \right\mid} \left(\frac{1}{x}\right) \cos \frac{\frac{1}{x}}{\left\mid \cos \right\mid} \left(\frac{1}{x}\right)$ .

Explanation:

$\frac{d}{\mathrm{dx}} \left(\arcsin u\right) = \frac{1}{\sqrt{1 - {u}^{2}}} \frac{\mathrm{du}}{\mathrm{dx}}$

So we get

$f ' \left(x\right) = \frac{1}{\sqrt{1 - {\sin}^{2} \left(\frac{1}{x}\right)}} \frac{d}{\mathrm{dx}} \left(\sqrt{{\sin}^{2} \left(\frac{1}{x}\right)}\right)$

$= \frac{1}{\sqrt{1 - {\sin}^{2} \left(\frac{1}{x}\right)}} \left[\frac{1}{2 \sqrt{{\sin}^{2} \left(\frac{1}{x}\right)}} \frac{d}{\mathrm{dx}} \left({\sin}^{2} \left(\frac{1}{x}\right)\right)\right]$

$= \frac{1}{2 \sqrt{{\sin}^{2} \left(\frac{1}{x}\right)} \left(2 \sqrt{1 - {\sin}^{2} \left(\frac{1}{x}\right)}\right)} \left[2 \sin \left(\frac{1}{x}\right) \cos \left(\frac{1}{x}\right) \left(- \frac{1}{x} ^ 2\right)\right]$

$= \frac{- \sin \left(\frac{1}{x}\right) \cos \left(\frac{1}{x}\right)}{{x}^{2} \sqrt{{\sin}^{2} \left(\frac{1}{x}\right)} \sqrt{1 - {\sin}^{2} \left(\frac{1}{x}\right)}}$

Because $1 - {\sin}^{2} u = {\cos}^{2} u$, we can write this as

$= \frac{- \sin \left(\frac{1}{x}\right) \cos \left(\frac{1}{x}\right)}{{x}^{2} \sqrt{{\sin}^{2} \left(\frac{1}{x}\right)} \sqrt{{\cos}^{2} \left(\frac{1}{x}\right)}}$.

Or, using $\sqrt{{u}^{2}} = \left\mid u \right\mid$, we might prefer

$= \frac{- \sin \left(\frac{1}{x}\right) \cos \left(\frac{1}{x}\right)}{{x}^{2} \left\mid \sin \left(\frac{1}{x}\right) \right\mid \left\mid {\cos}^{2} \left(\frac{1}{x}\right) \right\mid}$.

Those familiar with the fact that $\frac{u}{\left\mid u \right\mid} = \left\{\begin{matrix}1 & \text{if" & u > 0 \\ -1 & "if} & u < 0\end{matrix}\right.$,

the result might be clearest written as:

$- \frac{1}{x} ^ 2 \sin \frac{\frac{1}{x}}{\left\mid \sin \right\mid} \left(\frac{1}{x}\right) \cos \frac{\frac{1}{x}}{\left\mid \cos \right\mid} \left(\frac{1}{x}\right)$ .

In other words, the derivative at $x$ is

$- \frac{1}{x} ^ 2$ if $\sin \left(\frac{1}{x}\right)$ and $\cos \left(\frac{1}{x}\right)$ have the same sign,

and $\frac{1}{x} ^ 2$ if they have opposite signs. And, finally,

if either $\sin \left(\frac{1}{x}\right) = 0$ or $\cos \left(\frac{1}{x}\right) = 0$, then the derivative does not exist.