How do you differentiate #f(x)=sqrtsin(2-4x) # using the chain rule?

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Answer 1 · 2015-12-25T13:05:47

Chain rule states that #(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)#

In this case, we can rename #y=sqrt(u)#, #u=sinv# and #v=2-4x#, as we cannot derivate directly nor #sqrt(sin(f(x))# neither #sin(f(x))#.

So, step-by-step solution:

#(dy)/(du)=1/(2u^(1/2))#

#(du)/(dv)=cosv#

#(dv)/(dx)=-4#

#(dy)/(dx)=1/(2u^(1/2))cosv(-4)#

Substituting #u#:

#(dy)/(dx)=1/(2(sinv))^(1/2)cosv(-4)#

Substituting #v#:

#(dy)/(dx)=1/(2(sin(2-4x))^(1/2))cos(2-4x)(-4)#

#(dy)/(dx)=-(2cos(2-4x))/sqrt(sin(2-4x))#