Question

Диференціювання функцій
2 завдання
[8.6]
[8.22]
80 балів
СРОЧНО!

Answer 1

$\displaystyle\bf\\8.6\\\\y=(x^{2} -3x+3)(x^{2} +2x-1)\\\\y'=(x^{2} -3x+3)'\cdot(x^{2} +2x-1)+(x^{2} -3x+3)\cdot(x^{2} +2x-1)'=\\\\=(2x-3)\cdot(x^{2} +2x-1)+(x^{2} -3x+3)\cdot(2x+2)=\\\\=2x^{3} +4x^{2} -2x-3x^{2} -6x+3+2x^{3} +2x^{2} -6x^{2} -6x+6x+6=\\\\=4x^{3} -3x^{2} -8x+9\\\\\\8.22\\\\y=\Bigg(\frac{1+x^{2} }{1+x} \Bigg)^{5}$

$\displaystyle\bf\\y'=5\cdot\Bigg(\frac{1+x^{2} }{1+x}\Bigg)^{4} \cdot\Bigg( \frac{1+x^{2} }{1+x} \Bigg)'=\\\\\\=5\cdot\Bigg(\frac{1+x^{2} }{1+x}\Bigg)^{4} \cdot\frac{(1+x^{2} )'\cdot(1+x)-(1+x^{2} )\cdot(1+x)'}{(1+x)^{2} } =\\\\\\=5\cdot\Bigg(\frac{1+x^{2} }{1+x}\Bigg)^{4} \cdot\frac{2x\cdot(1+x)-(1+x^{2} )\cdot 1}{(1+x)^{2} } =\\\\\\=5\cdot\Bigg(\frac{1+x^{2} }{1+x}\Bigg)^{4} \cdot\frac{2x+2x^{2} -1-x^{2} }{(1+x)^{2} } =$

$\displaystyle\bf\\=5\cdot\Bigg(\frac{1+x^{2} }{1+x}\Bigg)^{4} \cdot\frac{x^{2} +2x-1}{(1+x)^{2} } =5\cdot\Bigg(\frac{1+x^{2} }{1+x}\Bigg)^{4} \cdot\frac{(x-1)^{2} }{(1+x)^{2} }$