<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] $$\ y=x^3\sqrt x-\frac{1}{x^2\sqrt x}=x^{\frac{7}{2}}-x^{-\frac{5}{2}}$$
$$\ y'=\frac{7}{2}x^{\frac{5}{2}}+\frac{5}{2}x^{-\frac{3}{2}}=$$
$$=\frac{7}{2}x^2\sqrt x+\frac{5}{2}\cdot\frac{1}{x\sqrt x}$$
</s>