How do you find the derivative of #y=ln((x-1)/(x+1))^(1/3)#?
2 Answers
Nov 11, 2016
# dy/dx = 2/(3x^2-3)#
Explanation:
# y = ln((x-1)/(x-1))^(1/3) #
# :. y = 1/3ln((x-1)/(x+1)) #
# :. 3y = ln((x-1)/(x+1)) #
# :. e^(3y) = ((x-1)/(x+1)) # ..... [1]
Differentiating the LHS implicity and the RHS using the quotient rule gives:
# :. 3e^(3y)dy/dx = ( (x+1)(1) - (x-1)(1) )/(x+1)^2#
# :. 3e^(3y)dy/dx = ( x+1-x+1 )/(x+1)^2#
# :. 3e^(3y)dy/dx = 2/(x+1)^2#
# :. 3((x-1)/(x+1))dy/dx = 2/(x+1)^2# (using [1])
# :. dy/dx = 2/3*1/(x+1)^2*(x+1)/(x-1)#
# :. dy/dx = 2/3*1/(x+1)*1/(x-1)#
# :. dy/dx = 2/3*1/(x^2-1)#
Hence,
# dy/dx = 2/(3x^2-3)#
Nov 17, 2016
# = 1/3ln((x-1)/(x-1))#
# = 1/3[ln(x-1)-ln(x+1)]#
Recall that
#d/dx(ln(x-1)) = 1/(x-1)# and#d/dx(ln(x+1)) = 1/(x+1)#
So
# = 2/(3(x^2-1))#