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Chain rule states that #(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)#
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Quotient rule states that for #y=f(x)/g(x)#, then #y'=(f'g-fg')/g^2#
Renaming #u=sqrt(v)# and #v=(4x-4)/(6x+10)#, we have the new #f(x)=2ln(u)# and can now start, but let's do it step-by-step.
#(dy)/(du)=2/u#
#(du)/(dv)=1/(2sqrt(v))#
#(dv)/(dx)=((4)(6x+10)-(4x-4)(6))/(6x+10)^2=(24x+40-24x+24)/(6x+10)^2=64/(6x+10)^2#
Aggregating:
#(dy)/(dx)=128/(u*2sqrt(v)*(6x+10)^2)#
Substituting #u#:
#(dy)/(dx)=128/(sqrt(v)2sqrt(v)(6x+10)^2)=128/(2v*(6x+10)^2)#
Substituting #v# (Holy Jesus!):
#(dy)/(dx)=128/(2((4x-4)/(cancel(6x+10)))(6x+10)^(cancel(2)))#
#(dy)/(dx)=64/((4x-4)(6x+10))=64/(24x^2+16x-40)#
#(dy)/(dx)=8/(3x^2+2x-5)#
Now, the second derivative:
#(dy^2)/(d^2x)=((0)(3x^2+2x+5)-(8)(6x+2))/(3x^2+2x-5)^2=-(48x+16)/(3x^2+2x-5)^2#
