How do you differentiate #f(x) = sqrt(sin^3(4-x) # using the chain rule?

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Answer 1 · 2018-05-13T23:45:05

the answer
#dy/dx=[-3*sin^2(4-x)*cos(4-x)]/(2sqrt(sin^3(4-x)))#

show below

#y= sqrt(sin^3(4-x)#

let suppose:

#u=4-x#

#(du)/dx=-1#

#r=sin^3(u)#

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

#y=sqrtr#

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

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

#dy/dx=-1*3*sin^2u*cosu*1/(2sqrtr)#

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

#dy/dx=[-3*sin^2(4-x)*cos(4-x)]/(2sqrt(sin^3(4-x)))#