use the definition to calculate the derivative of the following function. then find the values of the…

Explanation:

Step1: Recall the definition of the derivative

The definition of the derivative of a function $y = p(\theta)$ is $p'(\theta)=\lim_{h\rightarrow0}\frac{p(\theta + h)-p(\theta)}{h}$. Given $p(\theta)=\sqrt{5\theta}$, then $p(\theta + h)=\sqrt{5(\theta + h)}$. So, $\frac{p(\theta + h)-p(\theta)}{h}=\frac{\sqrt{5(\theta + h)}-\sqrt{5\theta}}{h}$.

Step2: Rationalize the numerator

Multiply the numerator and denominator by $\sqrt{5(\theta + h)}+\sqrt{5\theta}$: [ \begin{align*} \frac{\sqrt{5(\theta + h)}-\sqrt{5\theta}}{h}\times\frac{\sqrt{5(\theta + h)}+\sqrt{5\theta}}{\sqrt{5(\theta + h)}+\sqrt{5\theta}}&=\frac{5(\theta + h)-5\theta}{h(\sqrt{5(\theta + h)}+\sqrt{5\theta})}\ &=\frac{5\theta+ 5h-5\theta}{h(\sqrt{5(\theta + h)}+\sqrt{5\theta})}\ &=\frac{5h}{h(\sqrt{5(\theta + h)}+\sqrt{5\theta})}\ &=\frac{5}{\sqrt{5(\theta + h)}+\sqrt{5\theta}} \end{align*} ]

Step3: Find the limit as $h\rightarrow0$

[ p'(\theta)=\lim_{h\rightarrow0}\frac{5}{\sqrt{5(\theta + h)}+\sqrt{5\theta}}=\frac{5}{2\sqrt{5\theta}} ]

Step4: Calculate $p'(1)$

Substitute $\theta = 1$ into $p'(\theta)$: $p'(1)=\frac{5}{2\sqrt{5\times1}}=\frac{\sqrt{5}}{2}$

Step5: Calculate $p'(5)$

Substitute $\theta = 5$ into $p'(\theta)$: $p'(5)=\frac{5}{2\sqrt{5\times5}}=\frac{1}{2}$

Step6: Calculate $p'(\frac{3}{5})$

Substitute $\theta=\frac{3}{5}$ into $p'(\theta)$: $p'(\frac{3}{5})=\frac{5}{2\sqrt{5\times\frac{3}{5}}}=\frac{5}{2\sqrt{3}}=\frac{5\sqrt{3}}{6}$

Answer:

$p'(\theta)=\frac{5}{2\sqrt{5\theta}}$, $p'(1)=\frac{\sqrt{5}}{2}$, $p'(5)=\frac{1}{2}$, $p'(\frac{3}{5})=\frac{5\sqrt{3}}{6}$