i figure this relates so much to options .. some of you must know this.. find the dervative.. of 3/(2x) i thought you bring factor so you don't have fractions, put everything to a power.. like sqrt(x) is x^1/2 i'm very good at much more complex questions relative to this.. a simple one is differentiate f(x) = x(x^2+3) =x^3+3x f'(x) =3x^2+3 thats simple.. so question is.. how to find the derivative of 3/(2x) i'm using parenthesis here so you don't think its three halfs of x .. its three over two times x..
dx/dy for a quotient = ((derivitive of the numerator)(denominator)-(numerator)(derivitive of the denominator))/ (denominator)^2
because 3/(2x) = 3/2 *1/x there for lifting the x from the denominator you get 3/2(x^-1) then you differentiate -1 * (3/2) ... as the first part.. then subtract 1 from the exponent of x.. putting it at X^-2 so you end up with -3/2x^-2 that sure seems right.. and follows what you are saying..
you put x^-2 in denominator, which would switch it to numerator... 3/(2x) = (3/2) * x^-1 = -1x^-2 * (3/2) -(3x^-2)/2 or -3/(2x^2) http://www.wolframalpha.com/widgets/view.jsp?id=c44e503833b64e9f27197a484f4257c0
now.. (3/2)*((x^-2)/1) is the same as (3/2)*(1/(x^2)) yes moving the exponent up makes it negative and the reverse respectively.. such that negative three over 2x^2 is the same thing as 3/2( x ^-2)