Question

find the value of the derivative for the given function.
$left.\frac{dr}{d\theta}
ight|_{\theta = 1}$ if $r=\frac{5}{sqrt{17 - \theta}}$
$left.\frac{dr}{d\theta}
ight|_{\theta = 1}=$ 40 (type an integer or a simplified fraction.)

Explanation:

Step1: Rewrite the function

Rewrite $r = \frac{5}{\sqrt{17-\theta}}$ as $r = 5(17 - \theta)^{-\frac{1}{2}}$.

Step2: Apply the chain - rule

The derivative of $y = a\cdot u^n$ with respect to $x$ is $\frac{dy}{dx}=a\cdot n\cdot u^{n - 1}\cdot\frac{du}{dx}$. Here $a = 5$, $n=-\frac{1}{2}$, and $u = 17-\theta$. So $\frac{dr}{d\theta}=5\times(-\frac{1}{2})\times(17 - \theta)^{-\frac{1}{2}- 1}\times(-1)=\frac{5}{2}(17 - \theta)^{-\frac{3}{2}}$.

Step3: Evaluate the derivative at $\theta = 1$

Substitute $\theta = 1$ into $\frac{dr}{d\theta}$. We have $\frac{dr}{d\theta}\big|_{\theta = 1}=\frac{5}{2}(17 - 1)^{-\frac{3}{2}}=\frac{5}{2}\times16^{-\frac{3}{2}}=\frac{5}{2}\times\frac{1}{(16^{\frac{1}{2}})^3}=\frac{5}{2}\times\frac{1}{4^3}=\frac{5}{2}\times\frac{1}{64}=\frac{5}{128}$.

Answer:

$\frac{5}{128}$