What is the derivative of #ln((2x^2 - 3)/(2x^2 + 3))^(1/2)# ?

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Answer 1 · 2018-03-09T18:00:42

#y' = (12x)/(4x^4 - 9)#

#y = ln ((2x^2 - 3)/(2x^2 + 3))^(1/2)#

With the logarithm properties

#y = (1/2) ln ((2x^2 - 3)/(2x^2 + 3))#

Then

#y' = (1/2)* (1/((2x^2 - 3)/(2x^2 + 3)))*(((4x)(2x^2 + 3)-(2x^2 - 3)(4x))/(2x^2 + 3)^2)#

#y' = (1/2)* ((2x^2 + 3)/(2x^2 - 3))*((cancel(8x^3) + 12x-cancel(8x^3) + 12x)/(2x^2 + 3)^2)#

#y' = (1/2)* (cancel(2x^2 + 3)/(2x^2 - 3))*((24x)/cancel((2x^2 + 3)^2))#

#y' = (1/cancel(2))* (1/(2x^2 - 3))*((cancel(24)x)/(2x^2 + 3))#

#y' = (1)* (1/(2x^2 - 3))*((12x)/(2x^2 + 3))#

#y' = (12x)/((2x^2 - 3)(2x^2 + 3))#

#y' = (12x)/(4x^4 - 9)#