Question

How do you solve this question? Differentiate:

Answer 1

Answer:

$\displaystyle \frac{d}{dx}[x\sqrt{x}-x^{2}\sqrt{x}]$

$\displaystyle =\frac{3}{2}\sqrt{x} - \frac{5}{2}\;x\sqrt{x}$

Step-by-step explanation:

There's no need for the product rule. Consider the following:

• $\sqrt{x}$ is the same as $x^{1/2}$

As a result,

• $x\sqrt{x} = x\cdot x^{1/2} = x^{3/2}$, and
• $x^{2}\sqrt{x} = x^{2}\cdot x^{1/2} = x^{5/2}$.

$\displaystyle x\sqrt{x}-x^{2}\sqrt{x} = x^{3/2} - x^{5/2}$.

How to differentiate the first term, $x^{3/2}$?

Apply the power rule.

$\displaystyle \frac{d}{dx} [x^{3/2}] = \underbrace{3/2}_{\begin{aligned}&\text{from}\\[-0.5em]& \text{power}\end{aligned}} \;x^{(3/2) - 1} = \frac{3}{2}\;x^{1/2}$.

$\displaystyle \frac{d}{dx} [x^{5/2}] = \underbrace{5/2}_{\begin{aligned}&\text{from}\\[-0.5em]& \text{power}\end{aligned}} \;x^{(5/2) - 1} = \frac{5}{2}\;x^{3/2}$.

The derivative of difference is the difference of derivatives. Rewrite $x^{1/2}$ back as $\sqrt{x}$ since the question uses square roots rather than fraction power.

$\displaystyle\begin{aligned} \frac{d}{dx}[x\sqrt{x}-x^{2}\sqrt{x}] &=\frac{d}{dx}[x\sqrt{x}] - \frac{d}{dx}[x^{2}\sqrt{x}] \\&=\frac{3}{2}\;x^{1/2} - \frac{5}{2}\;x^{3/2}\\ &=\frac{3}{2} \sqrt{x} - \frac{5}{2} x\sqrt{x}\end{aligned}$.