330. Find $y^{\prime}$, if $y=\sqrt{x}+\cos ^{2} 3 x$.
Solution. $\quad y^{\prime}=\left(\sqrt{x}+\cos ^{2} 3 x\right)^{\prime}=(\sqrt{x})^{\prime}+\left(\cos ^{2} 3 x\right)^{\prime}=\frac{1}{2 \sqrt{x}}+$ $+2 \cos 3 x(-\sin 3 x) \cdot 3=\frac{1}{2 \sqrt{x}}-3 \sin 6 x$.

Here we used the fact that $(\sqrt{x})^{\prime}=\left(x^{1 / 2}\right)^{\prime}=\frac{1}{2} x^{-1 / 2}=$ $=\frac{1}{2 x^{1 / 2}}=\frac{1}{2 \sqrt{x}}$, and $2 \sin 3 x \cos 3 x=\sin 6 x$ (by the double-angle sine formula).

331-336. Find the derivatives of the following functions: