Question

Вычислите производные функций в точке x0= - 1 . a) y = 2x ^ 5 + x ^ 4 - x + 4 ; б) y = (x ^ 4 + x ^ 2)(x ^ 3 + x ^ 2) ;

в) y = 5x - 1/6x + 3(дробь)​

Answer 1

$\displaystyle\bf\\1)\\\\y=2x^{5} +x^{4} -x+4\\\\y'=2(x^{5} )'+(x^{4} )'-x'+4'=2\cdot 5x^{4} +4x^{3} -1+0=10x^{4} +4x^{3} -1\\\\y'(-1)=10\cdot(-1)^{4}+4\cdot(-1)^{3}-1=10-4-1=5\\\\\boxed{y'(-1)=5}\\\\\\2)\\\\y=(x^{4} +x^{2} )(x^{3} +x^{2} )\\\\y'=(x^{4}+x^{2} )'\cdot(x^{3} +x^{2} )+(x^{4} +x^{2} )\cdot(x^{3} +x^{2} )'=\\\\=(4x^{3} +2x)\cdot(x^{3} +x^{2} )+(x^{4} +x^{2} )\cdot(3x^{2} +2x)=\\\\=4x^{6} +4x^{5} +2x^{4} +2x^{3} +3x^{6}+2x^{5} +3x^{4} +2x^{3} =$

$\displaystyle\bf\\=7x^{6} +6x^{5} +5x^{4} +4x^{3} \\\\y'(-1)=7\cdot(-1)^{6} +6\cdot(-1)^{5} +5\cdot(-1)^{4}+4\cdot(-1)^{3} =7-6+5-4=2\\\\\boxed{y'(-1)=2}\\\\\\3)\\\\y=\frac{5x-1}{6x+3} \\\\\\y'=\frac{(5x-1)'\cdot(6x+3)-(5x-1)\cdot(6x+3)'}{(6x+3)^{2} } =\frac{5\cdot(6x+3)-6\cdot(5x-1)}{(6x+3)^{2} } \\\\\\=\frac{30x+15-30x+6}{(6x+3)^{2} } =\frac{21}{(6x+3)} \\\\\\y'(-1)=\frac{21}{[6\cdot(-1)+3]^{2} } =\frac{21}{(-3)^{2} } =\frac{21}{9} =2\frac{1}{3} \\\\\\\boxed{y'(-1)=2\frac{1}{3} }$