Find y' for y = 2 (sin x/2+ cos x/2)
Find $y'$ for $y = 2\!\left(\sin\dfrac{x}{2}+\cos\dfrac{x}{2}\right)$.
Explanation of Method
A half-angle inner function, where the factors cancel the outer constant.
Step 1: Differentiate each term by the chain rule
In both terms the inner function is $u = \dfrac{x}{2}$, with $u' = \dfrac{1}{2}$: \[\frac{d}{dx}\!\left(\sin\frac{x}{2}\right) = \cos\frac{x}{2}\cdot\frac{1}{2}, \qquad \frac{d}{dx}\!\left(\cos\frac{x}{2}\right) = -\sin\frac{x}{2}\cdot\frac{1}{2}.\]
Step 2: Combine
The outer factor $2$ cancels the two halves exactly: \[y' = 2\!\left(\frac{1}{2}\cos\frac{x}{2} - \frac{1}{2}\sin\frac{x}{2}\right) = \cos\frac{x}{2} - \sin\frac{x}{2}.\]
Step 3: Verification
\[x = 0:\quad y = 2(0+1) = 2,\quad y' = 1-0 = 1; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 1.0 \;\;\checkmark\]
\[y' = 0 \;\Longleftrightarrow\; \tan\frac{x}{2} = 1 \;\Longleftrightarrow\; x = \frac{\pi}{2}+2k\pi \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \cos\dfrac{x}{2} - \sin\dfrac{x}{2}\,}\]
Remark
Remark. Writing $y = 2\sqrt{2}\sin\!\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)$ shows the function is a single sinusoid of amplitude $2\sqrt{2}$ and period $4\pi$; its maximum occurs at $x = \tfrac{\pi}{2}$, which is where the derivative above vanishes.