Question

find the derivative of y=(sqrt(x^2-1))/(x-1)^(2/3). please help me with the calculations. it is so confusing.

Answer 1

use the formula either on D(uv)=uDv+vDu or D(u/v)=(vDu-uDv)/v^2
lets try the multiplication on derivative D(uv)=uDv+vDu, u=sqrt(x^2 -1) and v=(x-1)^(-2/3)
D(uv)=uDv+vDu=D[sqrt(x^2 -1)*(x-1)^(-2/3)]
=D(x^2 -1)^1/2 *(x-1)^(-2/3)
=(x^2 -1)^1/2 * D(x-1)^(-2/3) + (x-1)^(-2/3)*D(x^2 -1)^1/2
=(x^2 -1)^1/2*[(-2/3)(x-1)^(-5/3)] +(x-1)^(-2/3)*{(1/2)(x^2 -1)^(-1/2)2x
continue the algebraic process here

Answer 2

upon using the division rule of derivative D(u/v)=(vDu-uDv)/v^2, u=sqrt(x^2 -1) and v=(x-1)^(2/3)
D(u/v)=(vDu-uDv)/v^2
=[(x-1)^(2/3)* D(x^2 -1)^1/2 -sqrt(x^2 -1)*D(x-1)^(2/3) ]/[(x-1)^(2/3)]^2
=[(x-1)^(2/3) * (1/2)(x^2 -1)^(-1/2) 2x -sqrt(x^2 -1)*(2/3)(x-1)^(-1/3) ]/[(x-1)^(4/9)]
continue the algebraic process here and you will arrive at the same answer as in the multiplication rule

the answer is (x-2)/[3(x-1)^2/3 * sqrt(x^2 -1)]