How do you differentiate #sqrt(sin^3(1/x) # using the chain rule?

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Answer 1 · 2018-05-07T17:35:33

the answer
#dy/dx=[-3sin^2(1/x)*cos(1/x)]/[2*x^2*sqrt(sin^3(1/x))]#

Firstly
suppose:
#u=1/x#

#(du)/dx=-1/x^2#

#r=sin^3(u)#

#(dr)/(du)=3sin^2(u)*cos(u)#

#y=sqrtr#

#dy/(dr)=1/(2sqrtr)#

now

#dy/(dx)=dy/(dr)*(dr)/(du)(du)/dx#

#dy/dx=1/(2sqrtr)*3sin^2(u)*cos(u)[-1/x^2]#

#dy/dx=[-3sin^2(1/x)*cos(1/x)]/[2*x^2*sqrt(sin^3(1/x))]#