186. Using the definition of the derivative, find the derivative of the function $y=\sin 3x$ at the point $x=\pi / 9$.
Solution. Let's give some value $x$ an increment $\Delta x$; then the function $y$ will receive an increment

$$
\Delta y=\sin (3(x+\Delta x))-\sin 3 x=2 \cos \left(3 x+\frac{3 \Delta x}{2}\right) \sin \frac{3 \Delta x}{2}
$$

Form the ratio $\frac{\Delta y}{\Delta x}$ and take the limit as $\Delta x \rightarrow 0$:

$$
\begin{aligned}
& \lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{2 \cos \left(3 x+\frac{3 \Delta x}{2}\right) \cdot \sin \frac{3 \Delta x}{2}}{\Delta x}= \\
& =2 \lim _{\Delta x \rightarrow 0}\left(\cos \left(3 x+\frac{3 \Delta x}{2}\right) \frac{\sin \frac{3 \Delta x}{2}}{\Delta x}\right)=
\end{aligned}
$$

$$
=2 \lim _{\Delta x \rightarrow 0} \cos \left(3 x+\frac{3 \Delta x}{2}\right) \cdot \lim _{\Delta x \rightarrow 0} \frac{\sin \frac{3}{2} \Delta x}{\Delta x}=2 \cos 3 x \cdot \frac{3}{2}=3 \cos 3 x
$$

Thus, $(\sin 3 x)^{\prime}=3 \cos 3 x$.

Substituting the value of the argument $x=\pi / 9$, we get

$$
y^{\prime}(\pi / 9)=3 \cos \frac{\pi}{3}=3 \cdot \frac{1}{2}=\frac{3}{2}
$$