Find rho' with respect to phi for rho = (sin phi)/(1+ cos

Find $\rho'$ with respect to $\varphi$ for $\rho = \dfrac{\sin\varphi}{1+\cos\varphi}$.

Step 1: Identify the parts and apply the quotient rule

For $\rho = \dfrac{u}{v}$ recall $\rho' = \dfrac{u'v - uv'}{v^{2}}$. Here

\[ u = \sin\varphi,\quad u' = \cos\varphi, \qquad v = 1+\cos\varphi,\quad v' = -\sin\varphi. \]

Step 2: Substitute

Note that subtracting $uv'$ with $v' = -\sin\varphi$ adds a positive $\sin^{2}\varphi$: \[\rho' = \frac{\cos\varphi\,(1+\cos\varphi) - \sin\varphi\,(-\sin\varphi)}{(1+\cos\varphi)^{2}} = \frac{\cos\varphi + \cos^{2}\varphi + \sin^{2}\varphi}{(1+\cos\varphi)^{2}}.\]

Step 3: Apply the Pythagorean identity

Since $\sin^{2}\varphi + \cos^{2}\varphi = 1$, the numerator becomes $1+\cos\varphi$, which cancels one factor of the denominator: $\rho' = \frac{1+\cos\varphi}{(1+\cos\varphi)^{2}} = \frac{1}{1+\cos\varphi}$.

Step 4: Verification

The half-angle identities $\sin\varphi = 2\sin\tfrac{\varphi}{2}\cos\tfrac{\varphi}{2}$ and $1+\cos\varphi = 2\cos^{2}\tfrac{\varphi}{2}$ reveal what the function really is:

\[\rho = \frac{2\sin\frac{\varphi}{2}\cos\frac{\varphi}{2}}{2\cos^{2}\frac{\varphi}{2}} = \tan\frac{\varphi}{2}\]

\[\rho' = \frac{1}{2}\cdot\frac{1}{\cos^{2}\frac{\varphi}{2}} = \frac{1}{2\cos^{2}\frac{\varphi}{2}} = \frac{1}{1+\cos\varphi} \;\;\checkmark\]

\[\varphi = \frac{\pi}{2}:\quad \rho' = \frac{1}{1+0} = 1; \qquad \frac{\rho(1.5718)-\rho(1.5698)}{0.002} \approx 1.0 \;\;\checkmark\]