Find y' for y = sqrt(a²+b²-2ab cos phi), then evaluate y'

Find $y'$ for $y = \sqrt{a^{2}+b^{2}-2ab\cos\varphi}$, then evaluate $y'\!\left(\dfrac{\pi}{2}\right)$, $y'(\pi)$ and $y'\!\left(\dfrac{3\pi}{2}\right)$. Here $a, b \gt 0$.

Step 1: Apply the chain rule

Write $y = u^{1/2}$ with $u = a^{2}+b^{2}-2ab\cos\varphi$. Only the cosine depends on $\varphi$, so

\[ u' = -2ab\cdot(-\sin\varphi) = 2ab\sin\varphi, \]

and therefore

\[ y' = \frac{u'}{2\sqrt{u}} = \frac{2ab\sin\varphi}{2\sqrt{a^{2}+b^{2}-2ab\cos\varphi}} = \frac{ab\sin\varphi}{\sqrt{a^{2}+b^{2}-2ab\cos\varphi}}. \]

Step 2: Evaluate at $\varphi = \dfrac{\pi}{2}$

Here $\cos\tfrac{\pi}{2} = 0$ and $\sin\tfrac{\pi}{2} = 1$: \[y'\!\left(\frac{\pi}{2}\right) = \frac{ab\cdot 1}{\sqrt{a^{2}+b^{2}}} = \frac{ab}{\sqrt{a^{2}+b^{2}}}.\]

Step 3: Evaluate at $\varphi = \pi$

Here $\cos\pi = -1$ and $\sin\pi = 0$, and the radicand becomes a perfect square: \[y'(\pi) = \frac{ab\cdot 0}{\sqrt{a^{2}+b^{2}+2ab}} = \frac{0}{\sqrt{(a+b)^{2}}} = 0.\]

Step 4: Evaluate at $\varphi = \dfrac{3\pi}{2}$

Here $\cos\tfrac{3\pi}{2} = 0$ and $\sin\tfrac{3\pi}{2} = -1$: $y'\!\left(\frac{3\pi}{2}\right) = \frac{-ab}{\sqrt{a^{2}+b^{2}}}$.

Step 5: Verification

The expression is the law of cosines: $y$ is the third side of a triangle with sides $a$, $b$ and included angle $\varphi$.

\[\varphi = \pi:\quad y = \sqrt{(a+b)^{2}} = a+b, \text{ the degenerate triangle — the largest possible third side}\]

\[a=b=1,\ \varphi=\tfrac{\pi}{2}:\quad y' = \frac{1}{\sqrt2} \approx 0.70711; \qquad \frac{y(\tfrac{\pi}{2}+10^{-5})-y(\tfrac{\pi}{2}-10^{-5})}{2\times 10^{-5}} \approx 0.70711 \;\;\checkmark\]

That $y'(\pi) = 0$ is exactly right: at $\varphi = \pi$ the third side reaches its maximum $a+b$, so its rate of change vanishes.