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

2 Answers
Apr 14, 2016

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.)

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.

Apr 16, 2016

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)# .

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.