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

Using the chain rule, I get

So we get

the result might be clearest written as:

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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.)

Positive sine

Negative sine

#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)) :}#

#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.

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To differentiate ( \arcsin(\sqrt{\sin^2(\frac{1}{x})}) ) using the chain rule, follow these steps:

- Let ( u = \frac{1}{x} ).
- Find the derivative of ( u ) with respect to ( x ), which is ( \frac{du}{dx} = -\frac{1}{x^2} ).
- Now, let ( v = \sin(u) ).
- Find the derivative of ( v ) with respect to ( u ), which is ( \frac{dv}{du} = \cos(u) ).
- Then, let ( w = \sqrt{v^2} ).
- Find the derivative of ( w ) with respect to ( v ), which is ( \frac{dw}{dv} = \frac{v}{\sqrt{v^2}} ).
- Now, apply the chain rule to find ( \frac{dw}{du} ), which is ( \frac{dw}{du} = \frac{dw}{dv} \cdot \frac{dv}{du} ).
- Substitute ( v = \sin(u) ) and simplify.
- Finally, let ( z = \arcsin(w) ).
- Find the derivative of ( z ) with respect to ( w ), which is ( \frac{dz}{dw} = \frac{1}{\sqrt{1-w^2}} ).
- Apply the chain rule to find ( \frac{dz}{du} ), which is ( \frac{dz}{du} = \frac{dz}{dw} \cdot \frac{dw}{du} ).
- Substitute ( w = \sqrt{\sin^2(\frac{1}{x})} ) and simplify to obtain the final result.

The steps above involve several substitutions and derivatives using the chain rule, resulting in the differentiation of ( \arcsin(\sqrt{\sin^2(\frac{1}{x})}) ) with respect to ( x ).

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