Find y' for y = 1/(sin²x) + 1/(cos²x)

Find $y'$ for $y = \dfrac{1}{\sin^{2}x} + \dfrac{1}{\cos^{2}x}$.

Step 1: Rewrite as powers

\[ y = \bigl(\sin x\bigr)^{-2} + \bigl(\cos x\bigr)^{-2}. \]

Step 2: Differentiate by the chain rule

Note the sign reversal in the second term, where the inner derivative is $-\sin x$: \[\frac{d}{dx}\bigl(\sin x\bigr)^{-2} = -2\bigl(\sin x\bigr)^{-3}\cos x = -\frac{2\cos x}{\sin^{3}x},\]

\[ \frac{d}{dx}\bigl(\cos x\bigr)^{-2} = -2\bigl(\cos x\bigr)^{-3}\bigl(-\sin x\bigr) = \frac{2\sin x}{\cos^{3}x}. \]

Step 3: Combine over a common denominator

\[ y' = \frac{2\sin x}{\cos^{3}x} - \frac{2\cos x}{\sin^{3}x} = \frac{2\bigl(\sin^{4}x - \cos^{4}x\bigr)}{\sin^{3}x\cos^{3}x}. \]

Step 4: Apply the double-angle identities

The numerator factors as a difference of squares, and the denominator is a cube of a half: \[\sin^{4}x-\cos^{4}x = \bigl(\sin^{2}x-\cos^{2}x\bigr)\bigl(\sin^{2}x+\cos^{2}x\bigr) = -\cos 2x, \qquad \sin^{3}x\cos^{3}x = \frac{\sin^{3}2x}{8}.\]

\[ y' = \frac{2\bigl(-\cos 2x\bigr)}{\sin^{3}2x/8} = -\frac{16\cos 2x}{\sin^{3}2x}. \]

Step 5: Verification

\[x = \frac{\pi}{4}:\quad \cos 2x = 0 \;\Longrightarrow\; y' = 0, \text{ and } y\!\left(\tfrac{\pi}{4}\right) = 2+2 = 4 \text{ is the minimum} \;\;\checkmark\]

\[x = \frac{\pi}{6}:\quad y' = -\frac{16\cos\frac{\pi}{3}}{\sin^{3}\frac{\pi}{3}} = -\frac{8}{0.64952} \approx -12.31681 \;\;\checkmark\]