Question

Найти производную функции:
$y = ln(cos^{2}x + \sqrt{1 + cos^{4}x})$,
вводя промежуточное переменное $u = cos^{2}x$.

Answer 1

$\displaystyle y = ln(cos^{2}x+\sqrt{1+cos^{4}x})=ln(u+\sqrt{1+u^2});\\y'=\frac{(u+\sqrt{1+u^2})'}{u+\sqrt{1+u^2}}=\frac{u'+\frac{(1+u^2)'}{2\sqrt{1+u^2}} }{u+\sqrt{1+u^2}}=\frac{u'+\frac{2u'}{2\sqrt{1+u^2}} }{u+\sqrt{1+u^2}}=\\\frac{\frac{u'(2\sqrt{1+u^2})+2u'}{2\sqrt{1+u^2}} }{u+\sqrt{1+u^2}}=\frac{u'(2\sqrt{1+u^2})+2u'}{(u+\sqrt{1+u^2})2\sqrt{1+u^2}}=\\\frac{u'(2\sqrt{1+u^2}+2)}{2u\sqrt{1+u^2}+2+2u^2}$

$\displaystyle \frac{(cos^2x)'(\sqrt{1+cos^4x}+1)}{cos^2x\sqrt{1+cos^4x}+cos^4x+1}\\\frac{-2cosxsinx(\sqrt{1+cos^4x}+1)}{cos^2x\sqrt{1+cos^4x}+cos^4x+1}$