Find y' for y = (cos 2x+1)/(cos 2x-1)
Find $y'$ for $y = \dfrac{\cos 2x+1}{\cos 2x-1}$.
Explanation of Method
Half-angle identities turn a quotient into a single squared cotangent.
Step 1: Simplify with the double-angle identities
From $\cos 2x = 1-2\sin^{2}x = 2\cos^{2}x-1$ we obtain
\[ \cos 2x + 1 = 2\cos^{2}x, \qquad \cos 2x - 1 = -2\sin^{2}x. \]
Therefore the quotient is nothing but a squared cotangent: \[y = \frac{2\cos^{2}x}{-2\sin^{2}x} = -\frac{\cos^{2}x}{\sin^{2}x} = -\cot^{2}x.\]
Step 2: Differentiate by the chain rule
With $(\cot x)' = -\dfrac{1}{\sin^{2}x}$: \[y' = -2\cot x\cdot\bigl(\cot x\bigr)' = -2\cot x\cdot\left(-\frac{1}{\sin^{2}x}\right) = \frac{2\cot x}{\sin^{2}x}.\]
Step 3: Simplify
\[ y' = \frac{2\cos x}{\sin x}\cdot\frac{1}{\sin^{2}x} = \frac{2\cos x}{\sin^{3}x}. \]
Step 4: Verification
\[x = \frac{\pi}{4}:\quad \cos x = \sin x = \frac{\sqrt2}{2},\quad \sin^{3}x = \frac{\sqrt2}{4}\]
\[y' = \frac{2\cdot\frac{\sqrt2}{2}}{\frac{\sqrt2}{4}} = \frac{\sqrt2}{\frac{\sqrt2}{4}} = 4, \quad\text{and}\quad y\!\left(\frac{\pi}{4}\right) = -\cot^{2}\frac{\pi}{4} = -1\]
\[\frac{y(\pi/4+10^{-5})-y(\pi/4-10^{-5})}{2\times 10^{-5}} \approx 4.0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{2\cos x}{\sin^{3} x}, \qquad x \neq k\pi,\; k\in\mathbb{Z}\,}\]
Remark
Remark. Simplifying first turned a quotient-rule computation into a one-line chain rule. The excluded points $x = k\pi$ are exactly where $\cos 2x - 1 = 0$, so the two forms of the function have the same domain. Since $y = -\cot^{2}x \leq 0$, the function reaches its maximum $0$ at $x = \tfrac{\pi}{2}+k\pi$, which is where $\cos x$ — and hence $y'$ — vanishes.