Find the derivative of the following function f(x)=(x-3/2x+3)^2
Assuming the usual carelessness with parentheses, I gather you mean
f(x) = ((x-3)/(2x+3))^2
If f = u^2, f' = 2u u'
So, f' = 2(x-3)/(2x+3) * ((x-3)/(2x+3))'
= 2(x-3)/(2x+3) * [(2x+3)-2(x-3)/(2x+3)^2]
= 2(x-3)/(2x+3)*(9/(2x+3)^2)
= 18(x-3)/(2x+3)^3