Question

нужно найти 2ю производную

Answer 1

Ответ:

$\dfrac{6-2x^{2}}{x^{2}}-\dfrac{6+2x^{2}}{x^{2}}\ln x-2\ln^{2}x-4\ln x$

Пошаговое объяснение:

$y=(3-x^{2})\ln^{2}x;$

$y'=((3-x^{2})\ln^{2}x)'=(3-x^{2})' \cdot \ln^{2}x+(3-x^{2}) \cdot (\ln^{2}x)'=(3'-(x^{2})') \cdot \ln^{2}x+$

$+(3-x^{2}) \cdot 2\ln x \cdot \dfrac{1}{x}=(0-2x) \cdot \ln^{2}x+\dfrac{2}{x}(3-x^{2})\ln x=\dfrac{2}{x}(3-x^{2})\ln x-2x\ln^{2}x=$

$=\dfrac{6-2x^{2}}{x}\ln x-2x\ln^{2}x;$

$y''=(y')';$

$y''=\bigg (\dfrac{6-2x^{2}}{x}\ln x-2x\ln^{2}x \bigg )'=\bigg (\dfrac{6-2x^{2}}{x}\ln x \bigg )'-(2x\ln^{2}x)'=\bigg (\dfrac{6-2x^{2}}{x} \bigg )' \cdot$

$\cdot \ln x+\dfrac{6-2x^{2}}{x} \cdot (\ln x)'-((2x)' \cdot \ln^{2}x+2x \cdot (\ln^{2}x)')=\dfrac{(6-2x^{2})' \cdot x-(6-2x^{2}) \cdot x'}{x^{2}} \cdot$

$\cdot \ln x+\dfrac{6-2x^{2}}{x} \cdot \dfrac{1}{x}-(2\ln^{2}x+2x \cdot 2\ln x \cdot \dfrac{1}{x})=\dfrac{-4x \cdot x-(6-2x^{2})}{x^{2}} \cdot \ln x+\dfrac{6-2x^{2}}{x^{2}}-$

$-(2\ln^{2}x+4\ln x)=\dfrac{-4x^{2}+2x^{2}-6}{x^{2}} \cdot \ln x+\dfrac{6-2x^{2}}{x^{2}}-2\ln^{2}x-4\ln x=$

$=\dfrac{-2x^{2}-6}{x^{2}} \cdot \ln x+\dfrac{6-2x^{2}}{x^{2}}-2\ln^{2}x-4\ln x=\dfrac{6-2x^{2}}{x^{2}}-\dfrac{6+2x^{2}}{x^{2}}\ln x-2\ln^{2}x-$

$-4\ln x;$

$y''=\dfrac{6-2x^{2}}{x^{2}}-\dfrac{6+2x^{2}}{x^{2}}\ln x-2\ln^{2}x-4\ln x \ ;$