Question

Найти производную y=((2x^2)*sqrt(1+x^2))/3x^3

Answer 1

$\mathtt{f'(x)=[\frac{(2x^2-1)\sqrt{x^2+1}}{3x^3}]'=\frac{1}{3}[(2x^{-1}-x^{-3})\sqrt{x^2+1}]'=}\\\mathtt{\frac{1}{3}[(2x^{-1}-x^{-3})'\sqrt{x^2+1}+(2x^{-1}-x^{-3})(\sqrt{x^2+1})']=}\\\mathtt{\frac{1}{3}[(3x^{-4}-2x^{-2})\sqrt{x^2+1}+(2x^{-1}-x^{-3})*\frac{1}{2\sqrt{x^2+1}}*(x^2+1)']=}\\\mathtt{\frac{1}{3}[(3x^{-4}-2x^{-2})\sqrt{x^2+1}+\frac{2-x^{-2}}{\sqrt{x^2+1}}]=\frac{1}{3}*\frac{(3x^{-4}-2x^{-2})(x^2+1)+(2-x^{-2})}{\sqrt{x^2+1}}=}$$\mathtt{\frac{1}{3}*\frac{3x^{-2}+3x^{-4}-2-2x^{-2}+2-x^{-2}}{\sqrt{x^2+1}}=\frac{1}{3}*\frac{3x^{-4}}{\sqrt{x^2+1}}=\frac{x^{-4}}{\sqrt{x^2+1}}=\frac{1}{x^4\sqrt{x^2+1}}}$