Answer
$r' =\frac{1}{2\sqrt{\theta}}\sec\sqrt{\theta}\tan \sqrt{\theta}\tan\left(\frac{1}{\theta}\right)-\frac{1}{\theta^2}\sec\sqrt{\theta}\sec^2\left(\frac{1}{\theta}\right)$
Work Step by Step
We can use the product rule and the chain rule.
The given function is: $r=\sec\sqrt{\theta}\tan\left(\frac{1}{\theta}\right)$.
Let's denote $u =\sec\sqrt{\theta}$ and $v = \tan\left(\frac{1}{\theta}\right)$. Then, the function can be rewritten as: $r = uv$.
Now, let's find the derivative of $r$ with respect to $\theta$ using the product rule:
$r' = u'v + uv'$
First, let's find the derivative of $u$ with respect to $\theta$ using the chain rule: $$u' = \frac{1}{2\sqrt{\theta}}\sec\sqrt{\theta}\tan \sqrt{\theta}$$ Next, let's find the derivative of $v$ with respect to $\theta$ using the chain rule: $$v' = -\frac{1}{\theta^2}\sec^2\left(\frac{1}{\theta}\right)$$ Now, substitute the derivatives of $u$ and $v$ back into the product rule formula: $$\begin{aligned}
r' &=\frac{1}{2\sqrt{\theta}}\sec\sqrt{\theta}\tan \sqrt{\theta}\tan\left(\frac{1}{\theta}\right)\\
&-\frac{1}{\theta^2}\sec\sqrt{\theta}\sec^2\left(\frac{1}{\theta}\right).
\end{aligned}$$
