Question

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Answer 1

$\displaystyle\bf\\y=\frac{x^{3} +8}{2x+1} \\\\\\y'=\frac{(x^{3} +8)'\cdot(2x+1)-(x^{3} +8)\cdot(2x+1)'}{(2x+1)^{2} } =\\\\\\=\frac{3x^{2} \cdot(2x+1)-(x^{3} +8)\cdot2}{(2x+1)^{2} } =\frac{6x^{3} +3x^{2} -2x^{3}-16 }{(2x+1)^{2} } =\\\\\\=\frac{4x^{3} +3x^{2} -16}{(2x+1)^{2} } \\\\\\x_{0} =1\\\\\\y'(x_{0} )=y'(1)=\frac{4\cdot 1^{3} +3\cdot1^{2} -16}{(2\cdot 1+1)^{2} } =\frac{4+3-16}{3^{2} } =-\frac{9}{9} =-1\\\\\\Otvet \ : \ a. \ \ -1$