Question

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

Answer 1

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) = absu = {(u,"if",u >= 0),(-u,"if",u < 0) :}`, to get

`f(x) = arcsin(sqrt(sin^2(1/x))) = arcsin(abs(sin(1/x)))`

`= {(arcsin(sin(1/x)),"if",sin(1/x) >= 0),(arcsin(-sin(1/x)),"if",sin(1/x) < 0) :}`

`= {(arcsin(sin(1/x)),"if",sin(1/x) >= 0),(arcsin(sin(-1/x)),"if",sin(1/x) < 0) :}`

Now we have to find `arcsin(sin(1/x))` and `arcsin(sin(-1/x))` for the two cases `sin(1/x) >= 0` and `sin(1/x)< 0`

Positive sine

For every real number `x` with `sin(1/x) >= 0`, we know `1/x` is in Quadrant I or II.
So, there is exactly one integer `k` such that `1/x` lies in exactly one of the intervals below.

If `1/x in [2pik,pi/2+2pik)`, then `arcsin(sin(1/x))=1/x- 2pik`.

If `1/x in [pi/2+2pik, pi +2pik)`, then `arcsin(sin(1/x))=pi+2pik - 1/x`.

Negative sine

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

If `1/x in [-pi+2pik,-pi/2+2pik)`, then `-1/x` is in Quadrant II.
And, `-1/x in [pi/2-2pik,pi-2pik)` `arcsin(sin(-1/x))=pi-2pik - (-1/x)`.

If `1/x in [-pi/2+2pik, 2pik)`, then `-1/x` is in Quadrant I.
And, `-1/x in [-2pik,pi/2 - 2pik)` `arcsin(sin(-1/x))=(-1/x) + 2pik`.

Writing `f(x)`

#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(1/x) = 0 " 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.

Answer 2

Using the chain rule, I get `(-sin(1/x)cos(1/x))/(x^2sqrt(sin^2(1/x))sqrt(1-sin^2(1/x)))` which may be written: `-1/x^2 sin(1/x)/abssin(1/x) cos(1/x)/abscos(1/x)` .

Explanation:

`d/dx(arcsinu) = 1/sqrt(1-u^2) (du)/dx`

So we get

`f'(x) = 1/sqrt(1-sin^2(1/x)) d/dx(sqrt(sin^2(1/x)))`

`= 1/sqrt(1-sin^2(1/x)) [1/(2sqrt(sin^2(1/x))) d/dx(sin^2(1/x))]`

`= 1/(2sqrt(sin^2(1/x))(2sqrt(1-sin^2(1/x))))[2sin(1/x)cos(1/x) (-1/x^2)]`

`= (-sin(1/x)cos(1/x))/(x^2sqrt(sin^2(1/x))sqrt(1-sin^2(1/x)))`

Because `1-sin^2u = cos^2u`, we can write this as

`= (-sin(1/x)cos(1/x))/(x^2sqrt(sin^2(1/x))sqrt(cos^2(1/x)))`.

Or, using `sqrt(u^2) = absu`, we might prefer

`= (-sin(1/x)cos(1/x))/(x^2abs(sin(1/x))abs(cos^2(1/x)))`.

Those familiar with the fact that `u/absu = {(1,"if",u > 0),(-1,"if",u < 0):}`,

the result might be clearest written as:

`-1/x^2 sin(1/x)/abssin(1/x) cos(1/x)/abscos(1/x)` .

In other words, the derivative at `x` is

`-1/x^2` if `sin(1/x)` and `cos(1/x)` have the same sign,

and `1/x^2` if they have opposite signs. And, finally,

if either `sin(1/x)=0` or `cos(1/x)=0`, then the derivative does not exist.