Question

1. using the definition, find the derivative of the function. then determine the values of the derivative as instructed.

$p(x)=\sqrt{3x}$; $p(\frac{2}{3})$

Explanation:

Step1: Recall the definition of the derivative

The definition of the derivative of a function \(y = f(x)\) is \(f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}\). For \(p(x)=\sqrt{3x}\), we have \(p(x + h)=\sqrt{3(x + h)}\). Then \(\frac{p(x + h)-p(x)}{h}=\frac{\sqrt{3(x + h)}-\sqrt{3x}}{h}\).

Step2: Rationalize the numerator

Multiply the numerator and denominator by \(\sqrt{3(x + h)}+\sqrt{3x}\).

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Step3: Find the limit as \(h

ightarrow0\)
\(p^{\prime}(x)=\lim_{h
ightarrow0}\frac{3}{\sqrt{3(x + h)}+\sqrt{3x}}=\frac{3}{\sqrt{3x}+\sqrt{3x}}=\frac{3}{2\sqrt{3x}}=\frac{\sqrt{3}}{2\sqrt{x}}\)

Step4: Evaluate \(p^{\prime}(\frac{2}{3})\)

Substitute \(x = \frac{2}{3}\) into \(p^{\prime}(x)\). \(p^{\prime}(\frac{2}{3})=\frac{\sqrt{3}}{2\sqrt{\frac{2}{3}}}\). Simplify \(\frac{\sqrt{3}}{2\sqrt{\frac{2}{3}}}=\frac{\sqrt{3}\times\sqrt{3}}{2\sqrt{2}}=\frac{3}{2\sqrt{2}}=\frac{3\sqrt{2}}{4}\)

Answer:

\(\frac{3\sqrt{2}}{4}\)