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

1 Answer
Mar 9, 2018

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

Explanation:

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)