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

Question

Ezekiel Alexander · Accepted Answer

# dy/dx = 2/(3x^2-3)#

# 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)#

Scarlett Allison · Answer

# 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))#

Elijah Abram · Answer

undefined To find the derivative of y = ln((x - 1)/(x + 1))^(1/3), first use the chain rule. The derivative of ln(u) is (1/u) * u', where u = ((x - 1)/(x + 1))^(1/3). Then, find the derivative of u using the chain rule and simplify. The derivative of u is [(1/3) * ((x - 1)/(x + 1))^(-2/3) * ((1/(x + 1)) - (1/(x - 1)))]. Finally, multiply the derivatives of ln(u) and u together to get the derivative of y.