How do you find the derivative of #y=ln((x-1)/(x+1))^(1/3)#?

2 Answers
Steve M
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)#

Jim H
Nov 17, 2016

# y = ln((x-1)/(x-1))^(1/3) #

# = 1/3ln((x-1)/(x-1))#

# = 1/3[ln(x-1)-ln(x+1)]#

Recall that #d/dx(lnu) = 1/u (du)/dx#, so

#d/dx(ln(x-1)) = 1/(x-1)# and #d/dx(ln(x+1)) = 1/(x+1)#

So

#dy/dx = 1/3[1/(x-1)-1/(x+1)]#

# = 2/(3(x^2-1))#

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