How do you differentiate #f(x)=x/arcsinsqrt(ln(1/x^2)# using the chain rule?

1 Answer
Mar 31, 2018

Answer:

#dy/dx = {arcsin(sqrt{-2lnx})+ 1/{sqrt{-2lnx(1+2lnx))}}/(arcsin(sqrt{-2lnx}))^2#

Explanation:

Well... you definitely need the chain rule for this one!

I would recommend first splitting this up using the quotient rule.

#y=u/v to dy/dx = {v {du}/dx - u {dv}/dx}/v^2#

In this case #u=x# and #v=arcsin(sqrt{ln(1/x^2)})#

#{du}/dx = d/dx[x] = 1#

#{dv}/dx = d/dx[arcsin(sqrt{ln(1/x^2)})]#

This is where the chain rule comes in. Let's use the following substitutions:

#a=x^{-2}#
#b=ln(a)#
#c=b^{1/2}#

The chain rule states that #y=g(u)# where #u=f(x) to # #dy/dx=dy/{du} times {du}/dx#

Applying this we get

#{dv}/dx = d/dx[arcsin(sqrt{ln(1/x^2)})]=#

#d/{dc}[arcsin( c )] d/{db} [b^{1/2}]d/{da}[lna]d/dx[x^{-2}]#

This gives us

#1/{sqrt{1-c^2}} times 1/2 b^{-1/2} times1/a times (-2x^{-3})#

Now, putting in the values we get

#1/{1-(sqrt{ln(x^{-2})})^2} times 1/2(ln(x^{-2}))^{-1/2} times (x^{-2})^{-1} times -2x^{-3}#

This can be simplified to

#{dv}/dx=-1/{xsqrt{-2lnx(1+2lnx)#

Now we have the values we need for #dy/dx = {v {du}/dx - u {dv}/dx}/v^2#

Putting those in we get

#dy/dx = {arcsin(sqrt{-2lnx})+ 1/{sqrt{-2lnx(1+2lnx))}}/(arcsin(sqrt{-2lnx}))^2#