Find rho' with respect to phi for rho = (cos phi+ sin
Find $\rho'$ with respect to $\varphi$ for $\rho = \dfrac{\cos\varphi+\sin\varphi}{\cos\varphi-\sin\varphi}$.
Explanation of Method
A trigonometric quotient whose derivative collapses to a constant numerator.
Step 1: Identify the parts and apply the quotient rule
For $\rho = \dfrac{u}{v}$ recall $\rho' = \dfrac{u'v - uv'}{v^{2}}$. Here
\[ u = \cos\varphi+\sin\varphi,\quad u' = \cos\varphi-\sin\varphi, \qquad v = \cos\varphi-\sin\varphi,\quad v' = -\sin\varphi-\cos\varphi. \]
Notice that $u' = v$ and $v' = -u$: the two functions feed into each other.
Step 2: Substitute
\[ \rho' = \frac{(\cos\varphi-\sin\varphi)(\cos\varphi-\sin\varphi) - (\cos\varphi+\sin\varphi)\bigl(-\sin\varphi-\cos\varphi\bigr)}{(\cos\varphi-\sin\varphi)^{2}}. \]
Step 3: Simplify the numerator
The first product is a square; in the second, $-\bigl(-\sin\varphi-\cos\varphi\bigr) = \cos\varphi+\sin\varphi$, so it too becomes a square: \[\begin{aligned} (\cos\varphi-\sin\varphi)^{2} &= \cos^{2}\varphi - 2\sin\varphi\cos\varphi + \sin^{2}\varphi = 1 - \sin 2\varphi, \\[4pt] (\cos\varphi+\sin\varphi)^{2} &= \cos^{2}\varphi + 2\sin\varphi\cos\varphi + \sin^{2}\varphi = 1 + \sin 2\varphi. \end{aligned}\]
Adding them, the terms $\mp\sin 2\varphi$ cancel: $\bigl(1-\sin 2\varphi\bigr) + \bigl(1+\sin 2\varphi\bigr) = 2$.
Step 4: Write the final expression
\[ \rho' = \frac{2}{(\cos\varphi-\sin\varphi)^{2}} = \frac{2}{1-\sin 2\varphi}. \]
Step 5: Verification
Dividing numerator and denominator by $\cos\varphi$ reveals what the function really is: \[\rho = \frac{1+\tan\varphi}{1-\tan\varphi} = \frac{\tan\varphi + \tan\frac{\pi}{4}}{1 - \tan\varphi\tan\frac{\pi}{4}} = \tan\!\left(\varphi+\frac{\pi}{4}\right)\]
\[\rho' = \frac{1}{\cos^{2}\!\left(\varphi+\frac{\pi}{4}\right)} = \frac{2}{(\cos\varphi-\sin\varphi)^{2}} \;\;\checkmark\]
\[\varphi = 0:\quad \rho = \frac{1+0}{1-0} = 1, \qquad \rho' = \frac{2}{(1-0)^{2}} = 2 \;\;\checkmark\]
The last line uses $\cos\!\left(\varphi+\tfrac{\pi}{4}\right) = \dfrac{\cos\varphi-\sin\varphi}{\sqrt{2}}$, so that $\cos^{2}\!\left(\varphi+\tfrac{\pi}{4}\right) = \dfrac{(\cos\varphi-\sin\varphi)^{2}}{2}$.
Result
Final answer:
\[\boxed{\,\rho' = \dfrac{2}{(\cos\varphi-\sin\varphi)^{2}} = \dfrac{2}{1-\sin 2\varphi}\,}\]
Remark
Remark. Recognising $\rho = \tan\!\left(\varphi+\tfrac{\pi}{4}\right)$ explains everything at once: the derivative is positive wherever defined, and the excluded points $\varphi = \tfrac{\pi}{4}+k\pi$ — where $\cos\varphi = \sin\varphi$ — are exactly the vertical asymptotes of the shifted tangent.