Take the natural logarithm of both sides.
#lny= ln(((x^2 + 1)(sqrt(3x + 4)))/((2x - 3)sqrt(x^2 - 4)))#
Use the rules #ln(ab) = lna + lnb# and #ln(a/b) = lna - lnb#:
#lny = ln((x^2 + 1)sqrt(3x + 4)) - ln((2x - 3)sqrt(x^2 - 4))#
#lny = ln(x^2 + 1) + ln(3x + 4)^(1/2) - (ln(2x - 3) + ln(x^2 - 4)^(1/2))#
Now use the rule #lnn^a = alnn# to simplify further.
#lny = ln(x^2 + 1) + 1/2ln(3x + 4) - ln(2x - 3) - 1/2ln(x^2 - 4)#
Now differentiate using the chain rule.
#1/y(dy/dx) = (2x)/(x^2 + 1) + 3/(2(3x + 4)) - 2/(2x - 3) - (2x)/(2(x^2 - 4))#
#1/y(dy/dx) = (2x)/(x^2 + 1) + 3/(2(3x + 4)) - 2/(2x - 3) - x/(x^2 - 4)#
#dy/dx = y((2x)/(x^2 + 1) + 3/(2(3x +4)) - 2/(2x - 3) - x/(x^2 - 4))#
#dy/dx= ((x^2 + 1)sqrt(3x + 4))/((2x - 3)(sqrt(x^2 - 4)))((2x)/(x^2 + 1) + 3/(2(3x + 4)) - 2/(2x- 3) - x/(x^2 - 4)#
This can probably be simplified further, but I'll leave those workings up to you:).
Hopefully this helps!
