Given that y=ln([2x-1/2x=1]^1/2) , show that dy/dx= (1/2x-1)-(1/2x+1)

y=ln([2x-1/2x=1]^1/2)- can be written as y= [0.5ln(2x-1)]-[0.5ln(2x+1)] due to laws of logs. Take first term -- (0.5ln(2x-1)) and substitute 2x-1 for u. so u=2x-1 and y=0.5lnu Now dy/du=1/2u and du/dx=2. to find dy/dx, times these 2 together - giving dy/dx=2/2u = 1/u = 1/2x-1 Doing the same method for the second term gives you that d/dx of 0.5ln(2x+1)is dy/dx= 1/2x+1 Therfore by subbing these back into the orignial equation,l the derivative of the enitre equation becomes dy/dx = (1/[2x-1])-(1/[2x+1])