Find y' for y = sin^4 x + cos^4 x

Find $y'$ for $y = \sin^{4} x + \cos^{4} x$.

Step 1: Differentiate term by term

By the chain rule applied to each fourth power: \[\frac{d}{dx}\bigl(\sin^{4}x\bigr) = 4\sin^{3}x\cos x, \qquad \frac{d}{dx}\bigl(\cos^{4}x\bigr) = 4\cos^{3}x\cdot(-\sin x) = -4\cos^{3}x\sin x.\]

Step 2: Combine and factor

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

Step 3: Apply the double-angle identities

Using $2\sin x\cos x = \sin 2x$ and $\cos^{2}x - \sin^{2}x = \cos 2x$, so that $\sin^{2}x-\cos^{2}x = -\cos 2x$: \[y' = 2\sin 2x\cdot\bigl(-\cos 2x\bigr) = -2\sin 2x\cos 2x = -\sin 4x.\]

Step 4: Verification

An independent route uses the identity $\sin^{4}x+\cos^{4}x = 1 - \tfrac12\sin^{2}2x = \tfrac34 + \tfrac14\cos 4x$: \[y' = \frac{d}{dx}\!\left(\frac34 + \frac{\cos 4x}{4}\right) = \frac{-4\sin 4x}{4} = -\sin 4x \;\;\checkmark\]

\[x = 0:\quad y = 0+1 = 1,\quad y'(0) = -\sin 0 = 0, \text{ a maximum} \;\;\checkmark\]

\[x = \frac{\pi}{8}:\quad y' = -\sin\frac{\pi}{2} = -1; \qquad \frac{y(\pi/8+10^{-5})-y(\pi/8-10^{-5})}{2\times 10^{-5}} \approx -1.0 \;\;\checkmark\]