How do I find the derivative of ln*root3((x-1)/(x+1))ln⋅3√x−1x+1?
1 Answer
Nov 9, 2016
d/dx ln root(3)((x-1)/(x+1))= 2/(3x^2-3) ddxln3√x−1x+1=23x2−3
Explanation:
y = ln root(3)((x-1)/(x+1)) y=ln3√x−1x+1
:. y = ln ((x-1)/(x+1))^(1/3)
:. y = 1/3ln ((x-1)/(x+1))
:. y = 1/3{ln(x-1)-ln(x+1)} (law of logs)
Differentiating (using the chain rule) gives:
dy/dx = 1/3{1/(x-1)-1/(x+1)}
:. dy/dx = 1/3{ ( (x-1)-(x+1) )/((x-1)(x+1)) }
:. dy/dx = 1/3{ (-2)/(1-x^2) }
:. dy/dx = 2/3 1/(x^2-1)
:. dy/dx = 2/(3x^2-3)
