Answer:
[tex]\displaystyle \frac{d}{dx}[x\sqrt{x}-x^{2}\sqrt{x}][/tex]
[tex]\displaystyle =\frac{3}{2}\sqrt{x} - \frac{5}{2}\;x\sqrt{x}[/tex]
Step-by-step explanation:
There's no need for the product rule. Consider the following:
- [tex]\sqrt{x}[/tex] is the same as [tex]x^{1/2}[/tex]
As a result,
- [tex]x\sqrt{x} = x\cdot x^{1/2} = x^{3/2}[/tex], and
- [tex]x^{2}\sqrt{x} = x^{2}\cdot x^{1/2} = x^{5/2}[/tex].
[tex]\displaystyle x\sqrt{x}-x^{2}\sqrt{x} = x^{3/2} - x^{5/2}[/tex].
How to differentiate the first term, [tex]x^{3/2}[/tex]?
Apply the power rule.
[tex]\displaystyle \frac{d}{dx} [x^{3/2}] = \underbrace{3/2}_{\begin{aligned}&\text{from}\\[-0.5em]& \text{power}\end{aligned}} \;x^{(3/2) - 1} = \frac{3}{2}\;x^{1/2}[/tex].
[tex]\displaystyle \frac{d}{dx} [x^{5/2}] = \underbrace{5/2}_{\begin{aligned}&\text{from}\\[-0.5em]& \text{power}\end{aligned}} \;x^{(5/2) - 1} = \frac{5}{2}\;x^{3/2}[/tex].
The derivative of difference is the difference of derivatives. Rewrite [tex]x^{1/2}[/tex] back as [tex]\sqrt{x}[/tex] since the question uses square roots rather than fraction power.
[tex]\displaystyle\begin{aligned} \frac{d}{dx}[x\sqrt{x}-x^{2}\sqrt{x}] &=\frac{d}{dx}[x\sqrt{x}] - \frac{d}{dx}[x^{2}\sqrt{x}] \\&=\frac{3}{2}\;x^{1/2} - \frac{5}{2}\;x^{3/2}\\ &=\frac{3}{2} \sqrt{x} - \frac{5}{2} x\sqrt{x}\end{aligned}[/tex].
