# Math Help - Simplifying a Derivative

1. ## Simplifying a Derivative

I'm having an issue with the following derivative.

$y=\frac{x^3-3x^2+2\sqrt{x}-5}{\sqrt{x}}$

I use the product rule and get up to this point:

$=\frac{3x^2-6x+x^{-1/2}}{\sqrt{x}}-\frac{x^3-3x^2+2x^{1/2}-5}{2\sqrt{x^3}}$

Wolfram Alpha tells me the answer is:

$=\frac{5x^3-9x^2+5}{2\sqrt{x^3}}$

My fraction skills are not the best-- far from best. I tried multiplying the left side by $\frac{2\sqrt{x}}{2\sqrt{x}}$ but that made a big mess. Maybe it was the right thing to do but I just made a mess and couldn't see how to make it work. Thanks for any help. I've recently returned to school after a 17 year hiatus.

2. ## Re: Simplifying a Derivative

Hello, Maskawisewin!

$\text{Differentiate: }\:y\:=\:\frac{x^3-3x^2+2\sqrt{x}-5}{\sqrt{x}}$

I would not avoid the Quotient Formula by always changing a quotient to a product
. . unless you are very good at handling negative and fractional exponents.

Besides, I would handle it this way:

$y \;=\;\frac{x^3 - 3x^2 + 2x^{\frac{1}{2}} - 5}{x^{\frac{1}{2}}} \;=\;\frac{x^3}{x^{\frac{1}{2}}} - \frac{3x^2}{x^{\frac{1}{2}}} + \frac{2x^{\frac{1}{2}}}{x^{\frac{1}{2}}} - \frac{5}{x^{\frac{1}{2}}} \;=\; x^{\frac{5}{2}}\: \:- 3x^{\frac{3}{2}} \:+\: 2 \:-\: 5x^{-\frac{1}{2}}$

Then: . $y' \;=\;\tfrac{5}{2}x^{\frac{3}{2}} - \tfrac{9}{2}x^{\frac{1}{2}} + \tfrac{5}{2}x^{-\frac{3}{2}} \;=\;\tfrac{1}{2}x^{-\frac{3}{2}}\left(5x^3 - 9x^2 + 5\right)$

Therefore: . $y' \;=\;\frac{5x^3 - 9x^2 + 5}{2x^{\frac{3}{2}}}$

3. ## Re: Simplifying a Derivative

Originally Posted by Maskawisewin
I'm having an issue with the following derivative.

$y=\frac{x^3-3x^2+2\sqrt{x}-5}{\sqrt{x}}$

I use the product rule and get up to this point:

$=\frac{3x^2-6x+x^{-1/2}}{\sqrt{x}}-\frac{x^3-3x^2+2x^{1/2}-5}{2\sqrt{x^3}}$

Wolfram Alpha tells me the answer is:

$=\frac{5x^3-9x^2+5}{2\sqrt{x^3}}$

My fraction skills are not the best-- far from best. I tried multiplying the left side by $\frac{2\sqrt{x}}{2\sqrt{x}}$ but that made a big mess. Maybe it was the right thing to do but I just made a mess and couldn't see how to make it work. Thanks for any help. I've recently returned to school after a 17 year hiatus.
To get a common denominator, note that \displaystyle \begin{align*} 2\sqrt{x^3} = 2\sqrt{x\cdot x^2} = 2\sqrt{x}{\sqrt{x^2}} = 2x\sqrt{x} \end{align*}, so you will need to multiply the top and bottom of the first term by \displaystyle \begin{align*} 2x \end{align*}, not \displaystyle \begin{align*} 2\sqrt{x} \end{align*}.