Ответ:

По определению производной :  [tex]\bf y'=\lim\limits _{\Delta x \to 0}\,\dfrac{\Delta y}{\Delta x}[/tex]   .

[tex]\bf y=ctg\, 3x\\\\{\Delta y=y(x+{\Delta x)-y(x)=ctg(3x+3{\Delta x)-ctg3x=[/tex]  

[tex]\bf =\dfrac{cos(3x+3\Delta x)}{sin(3x+3\Delta x)}-\dfrac{cos3x}{sin3x}=\dfrac{cos(3x+3\Delta x)\ sin3x-cos3x\ sin(3x+3\Delta x)}{sin(3x+3\Delta x)\cdot sin3x}=\\\\\\=\dfrac{sin(3x-(3x+3\Delta x))}{sin(3x+3\Delta x)\cdot sin3x}=\dfrac{sin(-3\Delta x)}{sin(3x+3\Delta x)\cdot sin3x}=-\dfrac{sin(3\Delta x)}{sin(3x+3\Delta x)\cdot sin3x}[/tex]

[tex]\bf y'=\lim\limits _{\Delta x\to 0}\, \dfrac{\Delta y}{\Delta x}=\lim\limits _{\Delta x\to 0}\ \dfrac{-sin(3\Delta x)}{\Delta x\cdot sin(3x+3\Delta x)\cdot sin3x}=\Big[\ sin(3\Delta x)\sim 3\Delta x\ \Big]=\\\\\\=\lim\limits _{\Delta x\to 0}\, \dfrac{-3\, \Delta x}{\Delta x\cdot sin(3x+3\Delta x)\cdot sin3x}=\lim\limits _{\Delta x\to 0}\, \dfrac{-3}{sin(3x+\underbrace{\bf 3\Delta x}_{\to 0})\cdot sin3x}=\\\\\\=\lim\limits _{\Delta x\to 0}\, \dfrac{-3}{sin3x\cdot sin3x}=-\dfrac{3}{sin^23x}[/tex]