<s>[INST] derivative of $y=x^3\sqrt{x}-\frac{1}{x^2\sqrt{x}}$ $y=x^3\sqrt{x}-\dfrac{1}{x^2\sqrt{x}}$  
Both terms require the product rule, right? My try:  
$x^3\dfrac{1}{2}x^{-1/2}+3x^2x^{1/2}-\dfrac{-1}{2}x^{-3/2}x^{-2}--2x^{-3}x^{-1/2}$  
What am I doing wrong? The correct answer is:  $y\;'=3.5x^2\sqrt{x}+\dfrac{2.5}{x^3\sqrt{x}}$ and I don't see how what I got can reduce to this.
 [/INST] $$y=x^3x^{1/2} -x^{-2}x^{-1/2}$$
$$y'=3x^2x^{1/2} + \frac{1}{2}x^3x^{-1/2} -\left( -2 x^{-3}x^{-1/2} -\frac{1}{2}x^{-2}x^{-3/2}\right)$$
$$=3x^2x^{1/2} + \frac{1}{2}x^3x^{-1/2} +2 x^{-3}x^{-1/2} +\frac{1}{2}x^{-2}x^{-3/2}$$
$$=3x^{5/2} + \frac{1}{2}x^{5/2} + 2 x^{-7/2} + \frac{1}{2}x^{-7/2},$$
which matches the correct answer. You were not wrong in general (just a sign). You just needed to simplify a bit.
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