In each case, find $y'$
$\displaystyle{y = (x^2 - \frac{1}{x})(3x-4)}$
$\displaystyle{y = (\sqrt{x} + 5)(x^2 - 5)}$
$\displaystyle{y = (\sin x)(\cos x) - x \cos x + x^3}$
$\displaystyle{y = x^3 \sin x - x^3 \cos x}$
$\displaystyle{y = \ln e^x - e^x \ln x}$
Differentiate:
$\displaystyle{f\,(x) = (x^2+x)(3x+1)}$
$\displaystyle{f\,(x) = x^2e^x + \frac{1}{x^2}}$
$\displaystyle{f\,(x) = x^3\sin x + \sin (\frac{\pi}{4} )}$