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

Question

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

Emma Aldridge · Accepted Answer

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)# .

#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.

Luke Alvarez · Answer

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.)

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.

Christopher Allers · Answer

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

1. Let $ u = \frac{1}{x} $.
2. Find the derivative of $ u $ with respect to $ x $, which is $ \frac{du}{dx} = -\frac{1}{x^2} $.
3. Now, let $ v = \sin(u) $.
4. Find the derivative of $ v $ with respect to $ u $, which is $ \frac{dv}{du} = \cos(u) $.
5. Then, let $ w = \sqrt{v^2} $.
6. Find the derivative of $ w $ with respect to $ v $, which is $ \frac{dw}{dv} = \frac{v}{\sqrt{v^2}} $.
7. Now, apply the chain rule to find $ \frac{dw}{du} $, which is $ \frac{dw}{du} = \frac{dw}{dv} \cdot \frac{dv}{du} $.
8. Substitute $ v = \sin(u) $ and simplify.
9. Finally, let $ z = \arcsin(w) $.
10. Find the derivative of $ z $ with respect to $ w $, which is $ \frac{dz}{dw} = \frac{1}{\sqrt{1-w^2}} $.
11. Apply the chain rule to find $ \frac{dz}{du} $, which is $ \frac{dz}{du} = \frac{dz}{dw} \cdot \frac{dw}{du} $.
12. 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 $.