How do I find the derivative of #ln*root3((x-1)/(x+1))#?
1 Answer
Nov 9, 2016
#d/dx ln root(3)((x-1)/(x+1))= 2/(3x^2-3) #
Explanation:
# y = ln root(3)((x-1)/(x+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) #
