Question

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

$p(\theta)=\sqrt{5\theta}$; $p(1)$, $p(5)$, $pleft(\frac{2}{5}
ight)$

$p(\theta)=\frac{5}{2}\theta^{-\frac{1}{2}}$

Explanation:

Step1: Recall derivative definition

The derivative of a function $y = f(x)$ using the definition is $f'(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. For $p(\theta)=\sqrt{5\theta}$, we have $p'(\theta)=\lim_{h
ightarrow0}\frac{\sqrt{5(\theta + h)}-\sqrt{5\theta}}{h}$. Multiply the numerator and denominator by $\sqrt{5(\theta + h)}+\sqrt{5\theta}$ to rationalize the numerator: $\lim_{h
ightarrow0}\frac{5(\theta + h)-5\theta}{h(\sqrt{5(\theta + h)}+\sqrt{5\theta})}=\lim_{h
ightarrow0}\frac{5h}{h(\sqrt{5(\theta + h)}+\sqrt{5\theta})}=\lim_{h
ightarrow0}\frac{5}{\sqrt{5(\theta + h)}+\sqrt{5\theta}}=\frac{5}{2\sqrt{5\theta}}=\frac{5}{2}\theta^{-\frac{1}{2}}$.

Step2: Find $p'(1)$

Substitute $\theta = 1$ into $p'(\theta)$. $p'(1)=\frac{5}{2}(1)^{-\frac{1}{2}}=\frac{5}{2}$.

Step3: Find $p'(5)$

Substitute $\theta = 5$ into $p'(\theta)$. $p'(5)=\frac{5}{2}(5)^{-\frac{1}{2}}=\frac{5}{2\sqrt{5}}=\frac{\sqrt{5}}{2}$.

Step4: Find $p'(\frac{2}{5})$

Substitute $\theta=\frac{2}{5}$ into $p'(\theta)$. $p'(\frac{2}{5})=\frac{5}{2}(\frac{2}{5})^{-\frac{1}{2}}=\frac{5}{2}\sqrt{\frac{5}{2}}=\frac{5\sqrt{10}}{4}$.

Answer:

$p'(1)=\frac{5}{2}$, $p'(5)=\frac{\sqrt{5}}{2}$, $p'(\frac{2}{5})=\frac{5\sqrt{10}}{4}$