329. Find the value of the arbitrary function $f(x)=\sin ^{4} x-$ $-\cos ^{4} x$ at $x=\pi / 12$.
Solution. First, we transform the given function:

$$
f(x)=\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{2} x-\cos ^{2} x\right)=-\cos 2 x
$$

(since $\sin ^{2} \alpha+\cos ^{2} \alpha=1, \cos ^{2} \alpha-\sin ^{2} \alpha=\cos 2 \alpha$). Using the rule for differentiating a composite function, we get

$$
f^{\prime}(x)=(-\cos 2 x)^{\prime}=-(\cos 2 x)^{\prime}=-(-\sin 2 x) \cdot(2 x)^{\prime}=2 \sin 2 x
$$

At $x=\frac{\pi}{12}$, we have $f^{\prime}\left(\frac{\pi}{12}\right)=2 \sin \frac{\pi}{6}=1$.