i am having pre cal woes
today was the first day of school and i cant remember how to do this home work
$\displaystyle 3/4(x^2+1)^(-1/4)-(1/2)(x^2+1)^(3/4)$
step by step would be awesome
indeed, you did do something wrong. if you factor out $\displaystyle (x^2 + 1)^{-1/4}$ from $\displaystyle (x^2 + 1)^{3/4}$ you are left with $\displaystyle (x^2 + 1)$ since $\displaystyle (x^2 + 1)^{3/4} = (x^2 + 1)(x^2 + 1)^{-1/4}$
you have to make sure that when you factor something out, that the powers of the base add up to the original power
$\displaystyle \frac 34(x^2 + 1)^{-1/4} - \frac 12 (x^2 + 1)^{3/4} = (x^2 + 1)^{-1/4} \bigg( \frac 34 - \frac 12 (x^2 + 1) \bigg)$ ........note that if you multiply this out, you get back the original. this was not true with what you wrote.
to make it look nicer, we can factor out a 1/4, to get:
$\displaystyle \frac 14 (x^2 + 1)^{-1/4} [3 - 2(x^2 + 1)] = \frac 14 (x^2 + 1)^{-1/4}(1 - 2x^2) = \frac 14 (1 - 2x^2)(x^2 + 1)^{-1/4}$
look carefully at what i did. now's your chance to ask questions