# 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.

• Group the powers of $x$ and then group the terms with equal powers.
– user65203
Commented Oct 31, 2014 at 16:45
• In addition to grouping powers of $x$, as Yves suggests, you need to correct the "$+-2$" to "$--2$" (or just "+2"). Commented Oct 31, 2014 at 16:52

You can simplify the equation to $y=x^{7/2}-x^{-5/2}$ then try to differentiate
$$\ y=x^3\sqrt x-\frac{1}{x^2\sqrt x}=x^{\frac{7}{2}}-x^{-\frac{5}{2}}$$ $$\ y'=\frac{7}{2}x^{\frac{5}{2}}+\frac{5}{2}x^{-\frac{3}{2}}=$$ $$=\frac{7}{2}x^2\sqrt x+\frac{5}{2}\cdot\frac{1}{x\sqrt x}$$
$$y=x^3x^{1/2} -x^{-2}x^{-1/2}$$ $$y'=3x^2x^{1/2} + \frac{1}{2}x^3x^{-1/2} -\left( -2 x^{-3}x^{-1/2} -\frac{1}{2}x^{-2}x^{-3/2}\right)$$ $$=3x^2x^{1/2} + \frac{1}{2}x^3x^{-1/2} +2 x^{-3}x^{-1/2} +\frac{1}{2}x^{-2}x^{-3/2}$$ $$=3x^{5/2} + \frac{1}{2}x^{5/2} + 2 x^{-7/2} + \frac{1}{2}x^{-7/2},$$ which matches the correct answer. You were not wrong in general (just a sign). You just needed to simplify a bit.