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Ответ

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Автор - Universalka

$1)f(x)=(x+1)^{3}(x+2)\\f'(x)=[(x+1)^{3}]'*(x+2)+(x+1)^{3}*(x+2)'=3(x+1)^{2}*(x+2)+(x+1)^{3}\\f'(1)=3(1+1)^{3}*(1+2)+(1+1)^{3}=3*8*3+8=80\\Otvet:boxed{80}$

$2)f(x)=frac{x^{2}-1 }{x^{2}+1}\\f'(x)=frac{(x^{2}-1)'*(x^{2}+1)-(x^{2}-1)*(x^{2}+1)'}{(x^{2}+1)^{2}}=frac{2x(x^{2}+1)-2x(x^{2}-1)}{(x^{2}+1)^{2}}=frac{2x^{3}+2x-2x^{3}+2x}{(x^{2}+1)^{2}}=frac{4x}{(x^{2}+1)^{2}} \\f'(1)=frac{4*1}{(1^{2}+1)^{2}}=frac{4}{4}=1\\Otvet:boxed{1}$

$3)f(x)=frac{2x^{2}}{1-7x}\\f'(x)=frac{(2x^{2})'*(1-7x)-2x^{2}*(1-7x)'}{(1-7x)^{2}}=frac{4x*(1-7x)-2x^{2}*(-7) }{(1-7x)^{2}}=frac{4x-28x^{2}+14x^{2}}{(1-7x)^{2}}=frac{4x-14x^{2}}{(1-7x)^{2}}\\f'(1)=frac{4*1-14*1^{2}}{(1-7*1)^{2}}=frac{4-14}{(-6)^{2}}=-frac{10}{36}=-frac{5}{18} \\Otvet:boxed{-frac{5}{18}}$

$4)(frac{2x^{3}+3x^{2}+1}{x-1})'=frac{(2x^{3}+3x^{2}+1)'*(x-1)-(2x^{3}+3x^{2}+1)*(x-1)' }{(x-1)^{2}}=frac{(6x^{2}+6x)*(x-1)-(2x^{3}+3x^{2}+1)}{(x-1)^{2}}=frac{6x^{3}-6x^{2}+6x^{2}-6x-2x^{3}-3x^{2}-1}{(x-1)^{2}}=frac{4x^{3}-3x^{2}-6x-1}{(x-1)^{2}}$