Найдите сложную производную функции. Пожалуйста, решите мне с 6-12.

[latex]6)\quad y=6\sqrt{5x+3}\; \; ,\; \; y'=\frac{6\cdot 5}{2\sqrt{5x+3}}\\\\7)\quaf y=\sqrt{\frac{x}{9}-14}\; ,\; \; y'=\frac{1}{2\sqrt{\frac{x}{9}-14}}\cdot \frac{1}{9}\\\\8)\quad y=sin(8x-\frac{\pi}{2})\; ,\; \; y'=8\cdot cos(8x- \frac{\pi}{2} )\\\\9)\quad y=6cos(7x+\pi)\; ,\; \; y'=-6sin(7x+\pi )\cdot 7\\\\10)\quad y=tg(4x-\frac{\pi}{3})\; ,\; \; y'=\frac{1}{cos^2(4x-\frac{\pi}{3})}\cdot 4[/latex] [latex]11)\quad y=2ctg( \frac{x}{5} +\frac{\pi}{4} )\; ,\; \; y'=-2\cdot \frac{1}{5}\cdot \frac{1}{sin^2(\frac{x}{5}+\frac{\pi}{4} )}[/latex] [latex]12)\quad y=6sin^3(8x+\frac{\pi}{5})\\\\y'=6\cdot 3sin^2(8x+ \frac{\pi}{5} )\cdot cos(8x+\frac{\pi}{5})\cdot 8[/latex] [latex]P.S.\; \; y=6sin^3(8x+\frac{\pi}{5})=6\cdot u^3\; ,\; u=sin(8x+\frac{\pi}{5})\\\\\star \; \; (6\cdot u^3)'=6\cdot (u^3)'=6\cdot 3u^2\cdot u'\\\\y'=6\cdot 3sin^2(8x+\frac{\pi}{5})\cdot (sin(8x+\frac{\pi}{5}))'\\\\\star \; \; (sinu)'=cosu\cdot u'\; ,\; \; u=8x+\frac{\pi}{5}\\\\y'=6\cdot 3sin^2(8x+\frac{\pi}{5})\cdot cos(8x+\frac{\pi}{5})\cdot (8x+\frac{\pi}{5})'\\\\\star \; \; (kx+b)'=(kx)'+b'=k\cdot x'+0=k\cdot 1=k\\\\y'=6\cdot 3sin^2(8x+\frac{\pi}{5})\cdot cos(8x+\frac{\pi}{5})\cdot 8[/latex]