Срочно помогите!!!!100 баллов!

№ 1. а) [latex]y'=2*4x^{3}+3x^{2}=8x^{3}+3x^{2}[/latex] б) [latex]y'=1+cosx[/latex] в) [latex]y'= \frac{1}{3}*(-3)*x^{-4}+ \frac{1}{2 \sqrt{x} } =- \frac{1}{x^{4}}+ \frac{1}{2 \sqrt{x} } [/latex] г) [latex]y'=ctgx-x* \frac{1}{sin^{2}x} [/latex] № 2. а) [latex]f(x)=-30x+6x^{3}+10-2x^{2}[/latex] [latex]f'(x)=-30+6*3x^{2}-2*2x=18x^{3}-4x-30=2*(9x^{3}-2x-15)[/latex] [latex]f'(1)=2*(9-2-15)=-16[/latex] б) [latex]f'(x)= \frac{2x*(x+5)-x^{2}}{(x+5)^{2}} =\frac{2x^{2}+10x-x^{2}}{(x+5)^{2}}=\frac{x^{2}+10x}{(x+5)^{2}}=\frac{x(x+10)}{(x+5)^{2}}[/latex] [latex]f'(-4)=\frac{-4(-4+10)}{(-4+5)^{2}}=-24[/latex] в) [latex]f'(x)= \frac{1}{2 \sqrt{3+0.5x} } *0.5= \frac{1}{4 \sqrt{3+0.5x} }[/latex] [latex]f'(2)= \frac{1}{4 \sqrt{3+0.5*2} }=\frac{1}{4 \sqrt{3+1} }= \frac{1}{8} [/latex] № 3. [latex]f'(x)=-2sinx+1=0[/latex] [latex]2sinx=1[/latex] [latex]sinx=0.5[/latex] [latex]x= \frac{ \pi }{6} +2 \pi k[/latex], k∈Z [latex]x= \frac{5 \pi }{6} +2 \pi k[/latex], k∈Z Или [latex]x=(-1)^{k} \frac{ \pi }{6} + \pi k[/latex], k∈Z № 4. [latex]f'(x)= \frac{2}{3}*3x^{2}-8 =2x^{2}-8=2*(x^{2}-4)[/latex] [latex]g'(x)= \frac{2}{2 \sqrt{x} } = \frac{1}{ \sqrt{x} } [/latex] [latex] \frac{f'(x)}{g'(x)}= \frac{2*(x^{2}-4)}{ \frac{1}{ \sqrt{x} }} =2 \sqrt{x} *(x^{2}-4)=0[/latex] [latex]2 \sqrt{x} *(x^{2}-4)=0[/latex] [latex]x_{1}=0[/latex] [latex]x_{2}=2[/latex] [latex]x_{3}=-2<0[/latex] - посторонний корень ОДЗ: x≥0 Ответ: 0; 2