Помогитееееееее пожалуйста!!! Даю 20 баллов.

$1.\\1)\;f(x)=2x^3-\frac1{x^2}\\f'(x)=6x^2+\frac2{x^3}\\\\2)\;f(x)=(4-x^2)\sin x\\f'(x)=-2x\sin x+(4-x^2)\cos x\\\\3)\;f(x)=\frac{x^2-6x}{x+2}\\f'(x)=\frac{(2x-6)(x+2)-(x^2-6x)}{(x+2)^2}=\frac{2x^2-2x-12-x^2+6x}{(x+2)^2}=\frac{x^2+4x-12}{(x+2)^2}\\\\4)\;f(x)=2\sin2x-3tgx\\f'(x)=4\cos2x-\frac3{\cos^2x}\\\\5)\;y=\frac95x^5-\frac13x^3+13x-\frac7{15}\\y'=9x^4-x^2+13$

$6)\;y=12x^3-e^x\\y'=36x^2-e^x\\\\7)\;y=x^7+3\sin x\\y'=7x^6+3\cos x\\\\8)\;f(x)=(-5x+11)^4\\f'(x)=4(-5x+11)^2\cdot(-5x+11)'=-20(-5x+11)^3=20(5x-11)^3\\\\9)\;y=e^{3x}-2\cos x\\y'=3e^{3x}+2\sin x\\\\10)\;y=x\cdot tg3x+2^x\\y'=tg3x+\frac{3x}{\cos^23x}+2^x\cdot\ln2$

0\\32x>24\\x>\frac34" alt="2.\\f(x)=\frac14\cos4x\\f'(x)=-\sin4x\\f'(x_0)=f'(\frac\pi{16})=-\sin\left(4\cdot\farc\pi{16}\right)=-\sin\frac\pi4=-\frac1{\sqrt2}\\\\3.\\f(x)=(x^2-2x-3)^2\\f'(x)=2(x^2-2x-3)\cdot(x^2-2x-3)'=2(x^2-2x-3)(2x-2)\\2(x^2-2x-3)(2x-2)=0\\4(x-3)(x+1)(x-1)=0\\x_1=3,\;x_2=-1,\;x_3=1\\\\4.\\g(x)=(3-4x)^2\\g'(x)=2(3-4x)\cdot(3-4x)'=-8(3-4x)=32x-24\\32x-24>0\\32x>24\\x>\frac34" align="absmiddle" class="latex-formula">