Find rho' with respect to phi for rho = a sqrt(cos 2
Find $\rho'$ with respect to $\varphi$ for $\rho = a\sqrt{\cos 2\varphi}$, where $a$ is a constant.
Explanation of Method
A square root of a double-angle cosine — the lemniscate in polar form.
Step 1: Apply the chain rule
Write $\rho = a\bigl(\cos 2\varphi\bigr)^{1/2}$. There are two nested layers: the square root and the doubled angle. The inner derivative is
\[ u = \cos 2\varphi, \qquad u' = -\sin 2\varphi\cdot 2 = -2\sin 2\varphi. \]
Step 2: Substitute
\[ \rho' = a\cdot\frac{1}{2}\bigl(\cos 2\varphi\bigr)^{-1/2}\cdot\bigl(-2\sin 2\varphi\bigr) = -\frac{a\sin 2\varphi}{\sqrt{\cos 2\varphi}}. \]
The factor $\tfrac12$ from the root and the factor $2$ from the doubled angle cancel exactly.
Step 3: Verification
\[\varphi = 0:\quad \rho = a,\quad \rho' = -\frac{a\cdot 0}{1} = 0, \text{ the maximum of } \rho \;\;\checkmark\]
\[a = 1,\ \varphi = \frac{\pi}{6}:\quad \rho' = -\frac{\sin(\pi/3)}{\sqrt{\cos(\pi/3)}} = -\frac{\sqrt3/2}{\sqrt{1/2}} \approx -1.22474\]
\[\frac{\rho(\pi/6+10^{-5})-\rho(\pi/6-10^{-5})}{2\times 10^{-5}} \approx -1.22474 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\rho' = -\dfrac{a\sin 2\varphi}{\sqrt{\cos 2\varphi}}, \qquad \cos 2\varphi > 0\,}\]
Remark
Remark. The curve $\rho^{2} = a^{2}\cos 2\varphi$ is Bernoulli's lemniscate. It exists only where $\cos 2\varphi \geq 0$, i.e. $|\varphi| \leq \tfrac{\pi}{4}$ modulo $\pi$, which is why the two loops have that angular extent. At $\varphi = \pm\tfrac{\pi}{4}$ the radius reaches zero and $\rho'$ becomes infinite — the curve meets the origin with a well-defined tangent line but an unbounded rate of change of $\rho$.