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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